Mathematics 265 Introduction to Calculus I

Study Guide :: Appendix A

Answers to Exercises

Note: Answers to the odd-numbered exercises from the textbook are provided in the Student Solution Manual.

Unit 2

1. At a given time, the number of children in a family may be either changing or constant; in all cases, however, it eventually becomes a constant.
2. The acceleration due to gravity is a changing variable as gravity is different at the poles than at the equator, and different on mountaintops than at sea level. However, we treat it as a constant variable with a value of 9.8 metres per second per second.
3. The number of days in a year is a changing variable, because of leap years.
4. Any symbol is adequate, say ${\displaystyle V\text{.}}$ We would say, “Let ${\displaystyle V}$ be the volume of an ice cube.”
5. Let ${\displaystyle T}$ be the temperature of a living human being. Given that clinical hypothermia (a fatal condition if not treated) occurs at 33°C and hyperpyrexia (also fatal if untreated) occurs at 41.1°C, we could say: range ${\displaystyle 31\le T<44}$ degrees Celsius. The range you estimate may vary substantially from that given.
6. For the distance ${\displaystyle d}$ at time ${\displaystyle t\text{,}}$ the relation ${\displaystyle D}$ is ${\displaystyle D=\left\{\left(t,d\right)|d\text{is the distance travelled at time }t\right\}\text{.}}$
7. ${\displaystyle T}$ is the set of all pairs ${\displaystyle \left(\left(b,h\right),A\right)}$ such that ${\displaystyle A}$ is the area of a triangle with base ${\displaystyle b}$ and height ${\displaystyle h\text{.}}$
8. ${\displaystyle A}$ is related by ${\displaystyle T}$ to the pair ${\displaystyle \left(b,h\right)}$ if and only if ${\displaystyle A}$ is the area of a triangle with base ${\displaystyle b}$ and height ${\displaystyle h\text{.}}$
9. For the weight ${\displaystyle w}$ of a person of age ${\displaystyle a\text{,}}$ we have that the relation ${\displaystyle A=\left\{\left(w,a\right)|w\text{is the weight of a person of age }a\right\}\text{.}}$
10. ${\displaystyle T}$ is the set of all pairs ${\displaystyle \left(A,y\right)}$ such that ${\displaystyle A}$ is your age in the year ${\displaystyle y\text{,}}$ or ${\displaystyle A}$ is related to ${\displaystyle y}$ by the relation ${\displaystyle T}$ if and only if ${\displaystyle A}$ is your age in the year ${\displaystyle y\text{.}}$
11. Any pair of the form ${\displaystyle \left(s,s^2\right)}$ is in ${\displaystyle M}$; any pair not of this form is not in ${\displaystyle M\text{.}}$
12. If you were 34 years old in the year 2000, the answer is yes; if not, the answer is no.
13. Yes, ${\displaystyle A\left(b,h\right)\text{.}}$
14. ${\displaystyle s\left(30\right)=50}$
15. 1. ${\displaystyle T\left(c\right)}$
    2. ${\displaystyle L\left(t\right)}$
    3. ${\displaystyle r\left(P\right)}$
16. Let ${\displaystyle L}$ be the length, ${\displaystyle W}$ the width and ${\displaystyle A}$ the area of the rectangle. Thus ${\displaystyle A\left(L\right)=L\left(10-L\right)=10L-L^2\text{.}}$
17. Let ${\displaystyle V}$ be the volume, ${\displaystyle L}$ be the length and ${\displaystyle S}$ be the surface area of the cube. Thus ${\displaystyle S\left(V\right)=6{\left(\sqrt[3]{V}\right)}^2=6V^{2∕3}\text{.}}$
18. Let ${\displaystyle x}$ be the length of each side of the box, ${\displaystyle h}$ its height and ${\displaystyle S}$ its surface area. Then, ${\displaystyle S\left(x\right)=x^2+4x\left(2∕x^2\right)=x^2+\left(8∕x\right)\text{.}}$
19. 1. Let ${\displaystyle C}$ be the cost of renting the car, and ${\displaystyle n}$ the number of kilometres driven. So, ${\displaystyle C\left(n\right)=50+0.43n}$
    2. ${\displaystyle D_C=\left[0,\infty \right)\text{,}}$ ${\displaystyle R_C=\left[50,\infty \right)\text{.}}$
    3. Taking the distance from Edmonton to Calgary as 294 km, we calculate ${\displaystyle C\left(n\right)=50+0.43\left(294\right)=176.42\text{.}}$ Adding ${\displaystyle 5\%}$ GST gives us ${\displaystyle \$185.24\text{.}}$
20. 1. 1. ${\displaystyle D_f=ℝ}$
       2. ${\displaystyle f\left(1\right)=2\text{,}}$ ${\displaystyle f\left(-1\right)=-1}$
    2. 1. ${\displaystyle D_f=ℝ}$
       2. ${\displaystyle f\left(1\right)=2\text{,}}$ ${\displaystyle f\left(-1\right)=4}$
    3. 1. ${\displaystyle D_f=ℝ}$
       2. ${\displaystyle f\left(1\right)=1\text{,}}$ ${\displaystyle f\left(-1\right)=1}$
    4. 1. ${\displaystyle D_f=ℝ}$
       2. ${\displaystyle f\left(1\right)=5\text{,}}$ ${\displaystyle f\left(-1\right)=-1\text{.}}$
21. 1. When the temperature increases from 70 to 80 the volume also increases; when the temperature decreases from 80 to 72, the volume also decreases.
    2. ${\displaystyle V\left(80\right)=21}$ m³
    3. ${\displaystyle V\left(73\right)}$ is between 17 m³ and 18 m³
22. 1. ${\displaystyle ℝ-\left\{1∕3\right\}}$ in interval notation ${\displaystyle \left(-\infty ,1∕3\right)\cup \left(1∕3,\infty \right)\text{.}}$
    2. ${\displaystyle ℝ-\left\{2,1\right\}=\left(-\infty ,1\right)\cup \left(1,2\right)\cup \left(2,\infty \right)\text{.}}$
    3. ${\displaystyle \left[0,\infty \right)\text{.}}$
    4. ${\displaystyle \left[0,4\right]\text{.}}$
    5. ${\displaystyle \left(-\infty ,0\right)\cup \left(5,\infty \right)}$
23. 1. ${\displaystyle D_R=\left(\infty ,-3\right)\cup \left(-3,0\right)\cup \left(0,\infty \right)\text{.}}$
    2. ${\displaystyle D_F=\left(-\infty ,\infty \right)=ℝ\text{.}}$
24. 1. ${\displaystyle R\left(3x+1\right)=\frac{9x^2+6x-8}{9x^2+15x+4}\text{.}}$
    2. ${\displaystyle F\left(3x+1\right)=\frac{4}{\sqrt{27x^2+18x+7}}\text{.}}$
25. ${\displaystyle h\left(x\right)=\frac{1}{\left(x-2\right)\left(x-5\right)}}$
26.  

    

27. 1. ${\displaystyle D_f=ℝ\text{.}}$
    2. ${\displaystyle f\left(3\right)=3\text{.}}$
28. 1. No.
    2. ${\displaystyle g\left(-2\right)=1\text{.}}$
    3. ${\displaystyle g\left(2\right)=2\text{.}}$
    4. ${\displaystyle D_g=ℝ-0,4.}$
29. The vertical lines ${\displaystyle x=-3\text{,}}$ ${\displaystyle x=2}$ and ${\displaystyle x=5}$ are vertical asymptotes.
30.  

    

31. 1.  

       

    2.  

       

    3.  

       

    4.  

       

32. 1. ${\displaystyle F\left(x\right)}$ is a shift down by 8 units of the function ${\displaystyle f\left(x\right)=x\text{.}}$

       

    2. ${\displaystyle F\left(x\right)}$ is a shift to the left by 4 units of the quadratic function ${\displaystyle f\left(x\right)=x^2\text{.}}$

       

    3. ${\displaystyle F\left(x\right)}$ is a vertical compression of ${\displaystyle \cos x}$ by a factor of 3.

       

    4. ${\displaystyle F\left(x\right)}$ is a vertical stretch of the constant function ${\displaystyle f\left(x\right)=4}$ by a factor of 5.

       

Unit 3

2.  

   

4. 1. ${\displaystyle \underset{x\to {\frac{\pi }{2}}^{+}}{\lim }\tan x=-\infty \ne \infty =\underset{x\to {\frac{\pi }{2}}^{-}}{\lim }\tan x}$
   2. ${\displaystyle \underset{x\to 0^{+}}{\lim }\cot x=\infty \ne -\infty \underset{x\to 0^{-}}{\lim }\cot x}$
   3. ${\displaystyle \underset{x\to 0^{+}}{\lim }\csc x=\infty \ne -\infty =\underset{x\to 0^{-}}{\lim }\csc x}$
5.  

   

7. The functions in numbers 3, 5 and 6, are polynomials; hence, they are continuous everywhere, and their limits are 59, 205 and 256, respectively.

   The function in 4 is rational and 2 is in its domain, so the limit is ${\displaystyle \frac{3}{4}\text{.}}$

   The functions in 7, 8 and 9 are the composition of functions: in 7 the function is continuous at 1 and the limit is ${\displaystyle \frac{1}{8}}$; in 8 the function is continuous at ${\displaystyle -2}$ and the limit is 16; in 9 the function is continuous at 4 and the limit is 0.

13. 1. ${\displaystyle \underset{x\to \pi ∕2^{+}}{\lim }\sec x=\underset{x\to \pi ∕2^{+}}{\lim }\frac{1}{\cos x}=-\infty }$ since ${\displaystyle \underset{x\to \pi ∕2^{+}}{\lim }\cos x=0}$ and ${\displaystyle \cos x<0}$ for ${\displaystyle x\to \pi ∕2^{+}\text{.}}$
    2. ${\displaystyle \underset{x\to 0^{-}}{\lim }\csc x=\underset{x\to 0^{-}}{\lim }\frac{1}{\sin x}=-\infty }$ since ${\displaystyle \underset{x\to 0^{-}}{\lim }\sin x=0}$ and ${\displaystyle \sin x<0}$ for ${\displaystyle x\to 0^{-}}$
14. ${\displaystyle \underset{x\to \pi ∕2^{+}}{\lim }\sec x=-\infty }$ by part (a) of Exercise 13, above. ${\displaystyle \underset{x\to \pi ∕2^{-}}{\lim }\sec x=\infty }$ since ${\displaystyle \underset{x\to \pi ∕2^{-}}{\lim }\cos x=0}$ and ${\displaystyle \cos x>0}$ for ${\displaystyle x\to \pi ∕2^{-}\text{.}}$ Hence, ${\displaystyle \underset{x\to \pi ∕2^{+}}{\lim }\sec x\ne \underset{x\to \pi ∕2^{-}}{\lim }\sec x\text{.}}$ The limit does not exist.
15. 1. ${\displaystyle \underset{x\to 0}{\lim }{\csc }^{2}x=\infty }$
    2. ${\displaystyle \underset{x\to 0^{-}}{\lim }\frac{1}{\sin x+\tan x}=-\infty }$
    3. ${\displaystyle \underset{x\to 0}{\lim }\frac{1}{x^2\cos x}=\infty }$
17. 1. ${\displaystyle \underset{x\to 4^{+}}{\lim }\frac{x^2+3x-1}{x-4}=\infty }$ since ${\displaystyle \underset{x\to 4^{+}}{\lim }{x}^2+3x-1=27>0}$ and ${\displaystyle \underset{x\to 4^{+}}{\lim }\frac{1}{x-4}=\infty \text{.}}$
    2. ${\displaystyle \underset{t\to 4^{+}}{\lim }\frac{\sin \left(t-4\right)+3t}{t^2-16}=\infty }$ since ${\displaystyle \underset{t\to 4^{+}}{\lim }\sin \left(t-4\right)+3t=12>0}$ and ${\displaystyle \underset{t\to 4^{+}}{\lim }\frac{1}{t^2-16}=\infty \text{.}}$
18. ${\displaystyle x=-1∕2}$
19. The limit would be ${\displaystyle \infty }$ since the function ${\displaystyle f\left(x\right)+c}$ increases without bound as ${\displaystyle x\to a\text{,}}$ regardless of the value of ${\displaystyle c\text{.}}$
20. The limit would be ${\displaystyle -\infty }$ since the function ${\displaystyle f\left(x\right)+g\left(x\right)}$ decreases without bound as ${\displaystyle g\left(x\right)}$ get close to ${\displaystyle c}$ a fixed number, while ${\displaystyle f\left(x\right)}$ decreases without bound.

21. The limit would be ${\displaystyle \underset{x\to 0^{+}}{\lim }\frac{1}{x^2}=\infty }$ since ${\displaystyle \underset{x\to 0^{+}}{\lim }{x}^2=0}$ and ${\displaystyle x^2\ge 0}$ for any ${\displaystyle x\text{.}}$

    The limit

    ${\displaystyle \underset{x\to 0^{+}}{\lim }-\frac{1}{x}=-\infty }$

    since ${\displaystyle \underset{x\to 0^{+}}{\lim }x=0}$ and ${\displaystyle -x<0}$ for ${\displaystyle x\to 0^{+}\text{.}}$

    The limit

    ${\displaystyle \underset{x\to 0^{+}}{\lim }\frac{1}{x^2}-\frac{1}{x}=\underset{x\to 0^{+}}{\lim }\frac{1-x}{x^2}=\infty ,}$

    since ${\displaystyle \underset{x\to 0^{+}}{\lim }1-x=1>0}$ and ${\displaystyle \underset{x\to 0^{+}}{\lim }\frac{1}{x^2}=\infty \text{.}}$

27. 1. Laws of Limits, part 3 of Theorem 3.23
    2. Example 3.61 and part (a) of Theorem 3.17
    3. Theorem 3.18
    4. part 3 of Law of Limits 3.23 and Example 3.60
31. 1. By part (a) of Theorem 3.29, the limit is ${\displaystyle \frac{4}{2}=2}$
    2. By part (b) of Theorem 3.29, the limit is 0.
    3. By part (c) of Theorem 3.29.
    4. By part (c) of Theorem 3.29.
32. ${\displaystyle \frac{1}{x}\to 0^{+}}$ as ${\displaystyle x\to \infty }$; hence, if ${\displaystyle u=\frac{1}{x}\text{,}}$ then ${\displaystyle \underset{x\to 0^{+}}{\lim }\cot \left(u\right)=\infty \text{.}}$
33. By Remark 2.
    1. ${\displaystyle 5x^2+4x-1>0}$ for ${\displaystyle x\to \infty }$
    2. ${\displaystyle 3-4x+6x^3<0}$ as ${\displaystyle x\to -\infty }$
    3. ${\displaystyle \left(3x-1\right)\left(5-x\right)=-3x^2+16x-5<0}$ for ${\displaystyle x\to \infty \text{.}}$

34. 1. If ${\displaystyle c=0\text{,}}$ then ${\displaystyle cf\left(x\right)=0}$ and ${\displaystyle \underset{x\to a}{\lim }cf\left(x\right)=0=cL\text{.}}$

       Let ${\displaystyle c\ne 0}$ and ${\displaystyle \epsilon >0\text{.}}$ Since

       ${\displaystyle \underset{x\to a}{\lim }f\left(x\right)=L\text{ for }\frac{\epsilon }{\left|c\right|}>0,}$

       there exists a ${\displaystyle \delta >0}$ such that

       ${\displaystyle |f\left(x\right)-L|<\frac{\epsilon }{\left|c\right|}\text{ for }0<|x-a|<\delta .}$

       Hence,

       ${\displaystyle |cf\left(x\right)-cL|=\left|c\right||f\left(x\right)-L|<\frac{\epsilon }{\left|c\right|}\left(\left|c\right|\right)=\epsilon \text{ for }0<|x-a|<\delta .}$

    2. Let ${\displaystyle \epsilon >0\text{.}}$ Since ${\displaystyle \underset{u\to b}{\lim }f\left(u\right)=L\text{,}}$ there exists a ${\displaystyle {\delta }_1>0}$ such that

       ${\displaystyle |f\left(u\right)-L|<\epsilon \text{ for any }0<|u-b|<{\delta }_1.}$ (5)

       Since

       ${\displaystyle \underset{x\to a}{\lim }g\left(x\right)=b\text{ for }{\delta }_1>0,}$

       there exists a ${\displaystyle {\delta }_2>0}$ such that

       ${\displaystyle |g\left(x\right)-b|<{\delta }_1\text{ for }0<|x-b|<{\delta }_2.}$

       Therefore if ${\displaystyle 0<|x-a|<{\delta }_2\text{,}}$ then ${\displaystyle |g\left(x\right)-b|<{\delta }_1\text{.}}$ Substituting ${\displaystyle g\left(x\right)}$ for ${\displaystyle u}$ in Equation 1 gives us ${\displaystyle |f\left(g\left(x\right)\right)-L|<\epsilon \text{.}}$

35. Let ${\displaystyle \epsilon >0\text{.}}$ Since ${\displaystyle \underset{x\to a}{\lim }f\left(x\right)=L}$ and ${\displaystyle \underset{x\to a}{\lim }f\left(x\right)=K}$ for ${\displaystyle \frac{\epsilon }{2}>0\text{,}}$ there exist ${\displaystyle {\delta }_1>0}$ and ${\displaystyle {\delta }_2>0}$ such that

    1. ${\displaystyle |f\left(x\right)-L|<\frac{\epsilon }{2}}$ for ${\displaystyle 0<|x-a|<{\delta }_1,}$ and
    2. ${\displaystyle |f\left(x\right)-K|<\frac{\epsilon }{2}}$ for ${\displaystyle 0<|x-a|<{\delta }_2\text{.}}$

    Hence, for ${\displaystyle \delta =\text{min}\left({\delta }_1,{\delta }_2\right)}$ and ${\displaystyle 0<|x-a|<\delta \text{,}}$ statements I and II, hold and

    ${\displaystyle \begin{array}{l}\left|K-L\right|\\ \phantom{k}=\left|f\left(x\right)-K+L-f\left(x\right)\right|\le \left|f\left(x\right)-K\right|+\left|f\left(x\right)-L\right|\\ \phantom{k}<\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon \text{.}\end{array}}$

    Therefore, ${\displaystyle |K-L|<\epsilon }$ for any ${\displaystyle \epsilon >0\text{,}}$ and we conclude that ${\displaystyle |K-L|=0\text{.}}$ Otherwise, if ${\displaystyle |K-L|>0}$ for ${\displaystyle \epsilon =\frac{|K-L|}{2}\text{,}}$ we find that

    ${\displaystyle |K-L|<\frac{|K-L|}{2},}$

    and this implies that ${\displaystyle 1<\frac{1}{2}\text{,}}$ a contradiction.

    So ${\displaystyle |K-L|=0\text{,}}$ and we conclude that ${\displaystyle K=L\text{.}}$

36. 1. Since ${\displaystyle g\left(x\right)<0}$ for ${\displaystyle x\to a^{+}\text{,}}$ there exists a ${\displaystyle {\delta }_1>0}$ such that

       ${\displaystyle g\left(x\right)<0\text{ for }x-a<{\delta }_1.}$ I(a)

       Since ${\displaystyle \underset{x\to a^{+}}{\lim }g\left(x\right)=0}$ for ${\displaystyle N<0\text{,}}$ we know that ${\displaystyle -N>0\text{,}}$ and for this positive number, there exists a ${\displaystyle {\delta }_2>0}$ such that

       ${\displaystyle \left|g\left(x\right)\right|<-N\text{for}0<x-a<{\delta }_2.}$ II(a)

       Hence, for ${\displaystyle \delta =\text{min}\left({\delta }_1,{\delta }_2\right)\text{,}}$ Statements I(a) and II(a) hold, and

       ${\displaystyle \left|g\left(x\right)\right|=-g\left(x\right)<-N\text{ for }0<x-a<\delta .}$

       We conclude that

       ${\displaystyle \frac{1}{g\left(x\right)}<N\text{ for }0<x-a<\delta .}$

    2. Since ${\displaystyle g\left(x\right)>0}$ for ${\displaystyle x\to a\text{,}}$ there exists a ${\displaystyle {\delta }_1>0}$ such that

       ${\displaystyle g\left(x\right)>0\text{ for }|x-a|<{\delta }_1.}$ I(b)

       Since ${\displaystyle \underset{x\to a}{\lim }g\left(x\right)=0}$ for ${\displaystyle M>0\text{,}}$ there exists a ${\displaystyle {\delta }_2>0}$ such that

       ${\displaystyle \left|g\left(x\right)\right|<M\text{ for }0<|x-a|<{\delta }_2.}$ II(b)

       Hence, for ${\displaystyle \delta =\text{min}\left({\delta }_1,{\delta }_2\right)\text{,}}$ Statements I(b) and II(b) hold, and

       ${\displaystyle \left|g\left(x\right)\right|=g\left(x\right)<M\text{ for }0<|x-a|<\delta .}$

       We conclude that

       ${\displaystyle \frac{1}{g\left(x\right)}>M\text{ for }0<|x-a|<\delta .}$

    3. Since ${\displaystyle \underset{x\to a}{\lim }f\left(x\right)=L<0}$ for ${\displaystyle -L∕2>0\text{,}}$ there exists a ${\displaystyle {\delta }_1>0}$ such that

       ${\displaystyle |f\left(x\right)-L|<-\frac{L}{2}\text{ for }0<|x-a|<{\delta }_1.}$ I(c)

       Since ${\displaystyle \underset{x\to a}{\lim }g\left(x\right)=\infty }$ for any ${\displaystyle N<0\text{,}}$ the number ${\displaystyle \frac{2N}{L}>0\text{,}}$ and there exists a ${\displaystyle {\delta }_2>0}$ such that

       ${\displaystyle g\left(x\right)>\frac{2N}{L}\text{for}0<|x-a|<{\delta }_2.}$ II(c)

       Hence, for ${\displaystyle \delta =\text{min}\left({\delta }_1,{\delta }_2\right)\text{,}}$ Statements I(c) and II(c) hold, and for ${\displaystyle 0<|x-a|<\delta }$ we find that

       ${\displaystyle \frac{L}{2}<f\left(x\right)-L<-\frac{L}{2}\text{ and }\frac{3L}{2}<f\left(x\right)<\frac{L}{2}<0.}$

       Since ${\displaystyle g\left(x\right)>\frac{2N}{L}}$ and ${\displaystyle f\left(x\right)<\frac{L}{2}<0\text{,}}$ we conclude that

       ${\displaystyle f\left(x\right)g\left(x\right)<\frac{L}{2}\frac{2N}{L}=N.}$

    4. Since ${\displaystyle \underset{x\to a}{\lim }f\left(x\right)=L>0}$ for ${\displaystyle \frac{L}{2}>0\text{,}}$ there exists a ${\displaystyle {\delta }_1>0}$ such that

       ${\displaystyle |f\left(x\right)-L|<\frac{L}{2}\text{for}0<|x-a|<{\delta }_1.}$ I(d)

       Since ${\displaystyle \underset{x\to a}{\lim }g\left(x\right)=-\infty }$ for any ${\displaystyle N<0\text{,}}$ the number ${\displaystyle \frac{2N}{L}<0\text{,}}$ and there exists a ${\displaystyle {\delta }_2>0}$ such that

       ${\displaystyle g\left(x\right)<\frac{2N}{L}\text{ for }0<|x-a|<{\delta }_2.}$ II(d)

       Hence, for ${\displaystyle \delta =min\left({\delta }_1,{\delta }_2\right)\text{,}}$ Statements I(d) and II(d) hold, and for ${\displaystyle 0<|x-a|<\delta \text{,}}$ we find that

       ${\displaystyle -\frac{L}{2}<f\left(x\right)-L<\frac{L}{2}\text{ and }0<\frac{L}{2}<f\left(x\right)<\frac{3L}{2}.}$

       Since ${\displaystyle g\left(x\right)<\frac{2N}{L}}$ and ${\displaystyle f\left(x\right)>\frac{L}{2}>0\text{,}}$ we conclude that

       ${\displaystyle f\left(x\right)g\left(x\right)<\frac{L}{2}\frac{2N}{L}=N.}$

37. 1. Since ${\displaystyle \underset{x\to a}{\lim }g\left(x\right)=\infty }$ for any ${\displaystyle N<0\text{,}}$ the number ${\displaystyle -N>0\text{,}}$ and there exists a ${\displaystyle \delta >0}$ such that ${\displaystyle g\left(x\right)>-N}$ for ${\displaystyle 0<|x-a|<\delta \text{.}}$

       Hence, ${\displaystyle -g\left(x\right)<N}$ for ${\displaystyle 0<|x-a|<\delta \text{.}}$

    2. Since ${\displaystyle g\left(x\right)\ge f\left(x\right)}$ for ${\displaystyle x\to a\text{,}}$ there exists a ${\displaystyle {\delta }_1>0}$ such that

       ${\displaystyle g\left(x\right)\ge f\left(x\right)\text{ for }|x-a|<{\delta }_1.}$ I(b)

       Since ${\displaystyle \underset{x\to a}{\lim }g\left(x\right)=-\infty }$ for any ${\displaystyle N<0\text{,}}$ there exists a ${\displaystyle {\delta }_2>0}$ such that

       ${\displaystyle g\left(x\right)<N\text{ for }0<|x-a|<{\delta }_2.}$ II(b)

       Hence, for ${\displaystyle \delta =\text{min}\left({\delta }_1,{\delta }_2\right)}$ Statements I(b) and II(b) hold, and

       ${\displaystyle f\left(x\right)\le g\left(x\right)<N\text{ for }0<|x-a|<\delta .}$

    3. Since ${\displaystyle \underset{x\to a}{\lim }g\left(x\right)=\infty }$ and ${\displaystyle \underset{x\to a}{\lim }f\left(x\right)=\infty }$ for any ${\displaystyle M>0\text{,}}$ there exist ${\displaystyle {\delta }_1>0}$ and ${\displaystyle {\delta }_2>0}$ such that

       ${\displaystyle f\left(x\right)>\frac{M}{2}\text{ for }0<|x-a|<{\delta }_1.}$ I(c)

       and

       ${\displaystyle g\left(x\right)>\frac{M}{2}\text{ for }0<|x-a|<{\delta }_2.}$ II(c)

       Hence, for ${\displaystyle \delta =min\left({\delta }_1,{\delta }_2\right)}$ both statements are true and

       ${\displaystyle f\left(x\right)+g\left(x\right)>\frac{M}{2}+\frac{M}{2}=M\text{ for }0<|x-a|<\delta .}$

    4. Let ${\displaystyle M_1}$ be a positive number such that ${\displaystyle M_1+L>0\text{.}}$ Since ${\displaystyle \underset{x\to a}{\lim }f\left(x\right)=L\text{,}}$ there exists a ${\displaystyle {\delta }_1>0}$ such that

       ${\displaystyle |f\left(x\right)-L|<M_1+L\text{ for }0<|x-a|<M_1+L.}$

       Hence,

       ${\displaystyle -\left(M_1+L\right)<f\left(x\right)-L<M_1+L,}$

       and adding ${\displaystyle L}$ we find that

       ${\displaystyle -M_1<f\left(x\right)<M_1+2L\text{ for }0<|x-a|<M_1+L.}$ I(d)

       Since ${\displaystyle \underset{x\to a}{\lim }g\left(x\right)=\infty }$ for any ${\displaystyle M>0\text{,}}$ the number ${\displaystyle M_1+M>0\text{,}}$ and there exists a ${\displaystyle {\delta }_2>0}$ such that

       ${\displaystyle g\left(x\right)>M_1+M\text{ for }0<|x-a|<{\delta }_2.}$ II(d)

       If ${\displaystyle \delta =\text{min}\left({\delta }_1,{\delta }_2\right)\text{,}}$ Statements I(d) and II(d) hold, and for ${\displaystyle 0<|x-a|<\delta }$ we conclude that

       ${\displaystyle f\left(x\right)+g\left(x\right)>-M_1+M_1+M=M.}$

Unit 4

2. 1. dollars/thousand of litres
   2. When the demand for gas is between 30 and 50 thousand litres the cost of gas increases at a rate of 24 cents per thousand litres. The slope of the line passing through (30, G(30) and (50, G(50)) is 0.24.

17. ${\displaystyle g^{\prime }\left(x\right)=4{\left(f\left(x\right)\right)}^3{f}^{\prime }\left(x\right)}$ and ${\displaystyle h^{\prime }\left(x\right)=5{\left(f\left(x\right)\right)}^4{f}^{\prime }\left(x\right)\text{.}}$

23. ${\displaystyle \frac{d}{dx}\frac{1}{\cos x}=-\frac{-\sin x}{{\cos }^{2}x}=\frac{\sin x}{\cos x}\frac{1}{\cos x}=\tan x\sec x\text{.}}$

    ${\displaystyle \frac{d}{dx}\frac{1}{\sin x}=-\frac{\cos x}{{\sin }^{2}x}=-\frac{\cos x}{\sin x}\frac{1}{\sin x}=-\cot x\csc x\text{.}}$

24. ${\displaystyle \begin{array}{rl}{\displaystyle \frac{d}{dx}}{\sin }^{4}x& ={\displaystyle \frac{d}{dx}}{\sin }^{3}x\sin x=3{\sin }^{2}x\sin x\cos x+{\sin }^{3}x\cos x\\ & ={\sin }^{3}x\left(3\cos x+\cos x\right)=4{\sin }^{3}x\cos x\text{.}\end{array}}$

    ${\displaystyle \frac{d}{dx}{\sin }^{5}x=5{\sin }^{4}x\cos x.}$

Unit 5

1. A polynomial function is continuous everywhere, so it cannot have a vertical asymptote.

2. 1. ${\displaystyle f\left(t\right)=\frac{1}{\sqrt{3-t}}}$ is not defined at ${\displaystyle t=3}$ and ${\displaystyle D_f=\left(-\infty ,3\right)\text{.}}$ The vertical line ${\displaystyle x=3}$ is an asymptote because

      ${\displaystyle \underset{x\to 3^{-}}{\lim }f\left(t\right)=\frac{1}{\sqrt{3-t}}=\infty ,}$

      since ${\displaystyle \underset{x\to 3^{-}}{\lim }f\left(t\right)=\sqrt{3-t}=0}$ and ${\displaystyle \sqrt{3-t}>0}$ for ${\displaystyle x\to 3^{-}\text{.}}$

   2. ${\displaystyle g\left(x\right)=\frac{3x^2+5x-2}{x^2-4}}$ is not defined at ${\displaystyle x=2}$ and ${\displaystyle x=-2}$ and ${\displaystyle D_g=ℝ-\left\{2,-2\right\}\text{.}}$

      ${\displaystyle \underset{x\to -2^{+}}{\lim }\frac{3x^2+5x-2}{x^2-4}=\underset{x\to -2^{+}}{\lim }\frac{\left(3x-1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\underset{x\to -2^{+}}{\lim }\frac{3x-1}{x-2}=\frac{7}{4}.}$

      ${\displaystyle \underset{x\to 2^{+}}{\lim }\frac{3x^2+5x-2}{x^2-4}=\underset{x\to 2^{+}}{\lim }\frac{\left(3x-1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\underset{x\to 2^{+}}{\lim }\frac{3x-1}{x-2}=\infty ,}$

      since ${\displaystyle \underset{x\to 2^{+}}{\lim }3x-1=5>0\text{,}}$ ${\displaystyle \underset{x\to 2^{+}}{\lim }x-2=0}$ and ${\displaystyle x-2>0}$ for ${\displaystyle x\to 2^{+}\text{.}}$

      Hence, the vertical line ${\displaystyle x=2}$ is the vertical asymptote.

3. 1. no ${\displaystyle x}$-intercepts
   2. ${\displaystyle \left(-2,0\right)}$ and ${\displaystyle \left(1∕3,0\right)}$
   3. no ${\displaystyle x}$ intercepts
4. 1. ${\displaystyle \left(0,\frac{1}{\sqrt{3}}\right)}$
   2. ${\displaystyle \left(0,\frac{1}{2}\right)}$
5. 1. even
   2. neither
   3. odd
   4. neither
6. 1. even
   2. odd
12. Domain is ${\displaystyle \left(-\infty ,-3\right]}$ and ${\displaystyle \left[3,\infty \right)\text{,}}$ and the function is decreasing on ${\displaystyle \left(-\infty ,-3\right]}$ and increasing on ${\displaystyle \left[3,\infty \right)\text{.}}$ Therefore, there are no extreme points.
16. ${\displaystyle \left(-1,1\right)\text{,}}$ ${\displaystyle \left(-\frac{1}{\sqrt{5}},\frac{61}{125}\right)\text{,}}$ ${\displaystyle \left(\frac{1}{\sqrt{5}},\frac{61}{125}\right)\text{,}}$ ${\displaystyle \left(1,6\right)\text{.}}$

17. Acceleration is at the maximum when ${\displaystyle x=0\text{,}}$ it is at the minimum when

    ${\displaystyle x=\sqrt{\frac{3}{5}}.}$

25. ${\displaystyle p^{\prime }\left(x\right)=3a{x}^2+2bx+c=0}$ only if ${\displaystyle 4b^2-4\left(3a\right)\left(c\right)=4\left(b^2-3ac\right)>0\text{.}}$ By the second derivative test, ${\displaystyle p\left(x\right)}$ has a critical number only if ${\displaystyle b^2-3ac>0\text{.}}$

Unit 6

8. ${\displaystyle \begin{array}{ll}{\displaystyle \int {\cos }^{2}xdx}\hfill & ={\displaystyle \int {\displaystyle \frac{1+\cos \left(2x\right)}{2}}dx={\displaystyle \frac{1}{2}}{\displaystyle \int 1+\cos \left(2x\right)dx}}\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}\left(x+{\displaystyle \frac{\sin \left(2x\right)}{2}}\right)+C\hfill \\ \hfill & ={\displaystyle \frac{2x+\sin \left(2x\right)}{4}}+C\hfill \end{array}}$
9. 1. ${\displaystyle f\left(x\right)=x^2+3\text{,}}$ ${\displaystyle r=4\text{.}}$
   2. ${\displaystyle f\left(x\right)=3x-2\text{,}}$ ${\displaystyle r=20\text{.}}$
   3. ${\displaystyle f\left(x\right)=1+x+2x^2\text{,}}$ ${\displaystyle r=-\frac{1}{2}\text{.}}$
   4. ${\displaystyle f\left(x\right)=x^2+1\text{,}}$ ${\displaystyle r=-2\text{.}}$
   5. ${\displaystyle f\left(x\right)=2y+1\text{,}}$ ${\displaystyle r=-5\text{.}}$
   6. ${\displaystyle f\left(x\right)=5t+4\text{,}}$ ${\displaystyle r=-2.7\text{.}}$
   7. ${\displaystyle f\left(x\right)=4-t\text{,}}$ ${\displaystyle r=\frac{1}{2}\text{.}}$
   8. ${\displaystyle f\left(x\right)=\sin x\text{,}}$ ${\displaystyle r=6\text{.}}$
   9. ${\displaystyle f\left(x\right)=1+z^3\text{,}}$ ${\displaystyle r=-\frac{1}{3}\text{.}}$
10. 1. $f(x) = x^2 + 3, f'(x) = 2x, r = 4$, hence \[\frac{(x^2 + 3)^5}{5} + C\]
    2. $f(x) = 3x - 2, f'(x) = 3, r = 20$, hence \[\frac{(3x - 2)^{21}}{3(21)} = \frac{(3x - 2)^{21}}{63} + C\]
    3. $f(x) = 1 + x + 2x^2, f'(x) = 1 + 4x, r = -1/2$, hence \[2(1 + x + 2x^2)^{1/2} = 2\sqrt{1 + x + 2x^2} + C\]
    4. $f(x) = x^2 + 1, f'(x) = 2x, r = -2$, hence \[\frac{(x^2 + 1)^{-1}}{-1(2)} = -\frac{1}{2(x^2 + 1)} + C\]
    5. $f(y) = 2y + 1, f'(y) = 2, r = -5$, hence \[\frac{3(2y + 1)^{-4}}{-4(2)} = -\frac{3}{8(2y + 1)^4} + C\]
    6. $f(t) = 5t + 4, f'(t) = 5, r = -27$, hence \[\frac{(5t + 4)^{-26}}{-26(5)} = -\frac{1}{130(5t + 4)^{26}} + C\]
    7. $f(t) = 4 - t, f'(t) = -1, r = 1/2$, hence \[\frac{2(4 - t)^{3/2}}{-(3)} = -\frac{2(4 - t)^{3/2}}{3} + C\]
    8. $f(\theta) = \sin \theta, f'(\theta) = \cos \theta, r = 6$, hence \[\frac{\sin^7\theta}{7} + C\]
    9. $f(z) = 1 + z^3, f'(z) = 3z^2, r = -1/3$, hence \[\frac{3(1 + z^3)^{2/3}}{2(3)} = \frac{\sqrt[3]{(1 + z^3)^2}}{2} + C\]

    Note: You may differentiate the answers given, to see whether they are correct or not.

11. Note that the constant of integration has been omitted.
    1. ${\displaystyle u=3x}$; ${\displaystyle du=3dx}$ ${\displaystyle \frac{1}{3}\int \cos udu=\frac{1}{3}\sin u=\frac{1}{3}\sin \left(3x\right)\text{.}}$
    2. ${\displaystyle u=4+x^2}$; ${\displaystyle du=2xdx}$ ${\displaystyle \frac{1}{2}\int u^{10}dx=\frac{1}{2}\frac{u^{11}}{11}=\frac{{\left(4+x^2\right)}^{1}1}{22}\text{.}}$
    3. ${\displaystyle u=x^3+1}$; ${\displaystyle du=3x^{2}dx}$ ${\displaystyle \frac{1}{3}\int u^{1∕2}du=\frac{2}{9}{u}^{3∕2}=\frac{2\sqrt{{\left(x^2+1\right)}^3}}{9}\text{.}}$
    4. ${\displaystyle u=\sqrt{x}}$; ${\displaystyle du=\frac{1}{2\sqrt{x}}dx}$ ${\displaystyle 2\int \sin udu=-2\cos u=-2\cos \sqrt{x}\text{.}}$
    5. ${\displaystyle u=1+2x}$; ${\displaystyle du=2dx}$ ${\displaystyle 2\int u^{-3}du=2\frac{u^{-2}}{-2}=-\frac{1}{{\left(1+2x\right)}^2}\text{.}}$
    6. ${\displaystyle u=\cos \theta }$; ${\displaystyle du=-\sin \theta d\theta }$ ${\displaystyle -\int u^{4}du=-\frac{u^5}{5}=-\frac{{\cos }^5\theta }{5}}$
    7. ${\displaystyle u=x^2+3}$; ${\displaystyle du=2xdx}$ ${\displaystyle \int u^{4}du=\frac{u^5}{5}=\frac{{\left(x^2+3\right)}^5}{5}}$
    8. ${\displaystyle u=x^3+5}$; ${\displaystyle du=3x^{2}dx}$ ${\displaystyle \frac{1}{3}\int u^{9}du=\frac{u^{10}}{30}=\frac{{\left(x^3+5\right)}^{10}}{30}\text{.}}$
    9. ${\displaystyle u=3x-2}$; ${\displaystyle du=3dx}$ ${\displaystyle \frac{1}{3}\int u^{20}du=\frac{u^{21}}{63}=\frac{{\left(3x-2\right)}^{21}}{63}\text{.}}$
    10. ${\displaystyle u=2-x}$; ${\displaystyle du=-dx}$ ${\displaystyle -\int u^{6}du=-\frac{u^7}{7}=-\frac{{\left(2-x\right)}^7}{7}\text{.}}$

    11. ${\displaystyle u=1+x+2x^2}$; ${\displaystyle du=\left(1+4x\right)dx}$;

        ${\displaystyle \int u^{-1∕2}du=2u^{1∕2}=2\sqrt{1+x+2x^2}\text{.}}$

    12. ${\displaystyle u=2y+1}$; ${\displaystyle du=2dy}$ ${\displaystyle \frac{3}{2}\int u^{-5}du=-\frac{3u^{-4}}{8}=-\frac{3}{8{\left(2y+1\right)}^4}\text{.}}$
    13. ${\displaystyle u=4-t}$; ${\displaystyle du=-dt}$ ${\displaystyle -\int u^{1∕2}du=-\left(\frac{2}{3}\right)u^{3∕2}=\frac{2{\left(4-t\right)}^{3∕2}}{3}\text{.}}$
    14. ${\displaystyle u=\pi t}$; ${\displaystyle du=\pi dt}$ ${\displaystyle \frac{1}{\pi }\int \sin udu=-\frac{\cos u}{\pi }=-\frac{\cos \left(\pi t\right)}{\pi }\text{.}}$
    15. ${\displaystyle u=\sqrt{t}}$; ${\displaystyle du=\frac{1}{2\sqrt{t}}dt}$ ${\displaystyle 2\int \cos udu=2\sin u=2\sin \sqrt{t}\text{.}}$
    16. ${\displaystyle u=\sin \theta }$; ${\displaystyle du=\cos \theta d\theta }$ ${\displaystyle \int u^{6}du=\frac{u^7}{7}=\frac{{\sin }^7\theta }{7}\text{.}}$
    17. ${\displaystyle u=1+z^3}$; ${\displaystyle du=3z^{2}dz}$ ${\displaystyle \frac{1}{3}\int u^{-1∕3}du=\frac{3u^{2∕3}}{6}=\frac{{\left(1+z^3\right)}^{2∕3}}{2}}$
    18. ${\displaystyle u=\cot x}$; ${\displaystyle du=-{\csc }^{2}xdx}$ ${\displaystyle -\int u^{1∕2}du=-\frac{2u^{3∕2}}{3}=-\frac{2{\cot }^{3∕2}x}{3}\text{.}}$
    19. ${\displaystyle u=\sec x}$; ${\displaystyle du=\sec \tan xdx}$ ${\displaystyle \int <!--\; \int \; -->udu=\frac{u^2}{2}=\frac{{\sec }^{2}x}{2}}$ also ${\displaystyle u=\tan x}$ and ${\displaystyle \int udu=\frac{u}{2}=\frac{\tan x}{2}\text{.}}$
12. Note that the constant of integration has been omitted.

    1. ${\displaystyle u=x+2}$; ${\displaystyle du=dx}$ and ${\displaystyle u-2=x}$

       ${\displaystyle \int \frac{u-2}{\sqrt[4]{u}}du=\frac{4u^{7∕4}}{7}-\frac{8u^{3∕4}}{3}=\frac{4{\left(x+2\right)}^{7∕4}}{7}-\frac{8{\left(x+2\right)}^{3∕4}}{3}}$

    2. ${\displaystyle u=x-1}$; ${\displaystyle du=dx}$ and ${\displaystyle x^2={\left(u+1\right)}^2}$

       ${\displaystyle \int \frac{{\left(u+1\right)}^2}{\sqrt{u}}du=\int u^{3∕2}+2u^{-1∕2}du=\frac{2u^{5∕2}}{5}+\frac{4u^{3∕2}}{3}+2u^{1∕2}+C}$

       ${\displaystyle =\frac{2}{5}{\left(x-1\right)}^{5∕2}+\frac{4}{3}{\left(x-1\right)}^{3∕2}+2\sqrt{x-1}+C}$

13. Note that the constant of integration has been omitted.

    ${\displaystyle u=x^2-3}$; ${\displaystyle du=2xdx}$ and ${\displaystyle x^2=u+3}$

    ${\displaystyle \begin{array}{l}{\displaystyle \int \frac{x^3}{\sqrt{x^2-3}}}dx=\frac{1}{2}{\displaystyle \int u^{-1/2}\left(u+3\right)du=\frac{1}{2}{\displaystyle \int u^{1/2}+3u^{-1/2}}}\\ \phantom{=}=\frac{1}{2}\left[\frac{2}{3}{u}^{3/2}+6u^{1/2}\right]=\frac{{\left(x^2-3\right)}^{3/2}}{3}+3\sqrt{x^2-3}=\frac{\sqrt{x^2-3}\left(x^2+6\right)}{3}\text{.}\end{array}}$

16. Assume that the equation has two roots $r_1$, $r_2$, hence the function $f(x)=1+2x+x^3+4x^5$ has these zero roots. Hence $f(r_1)=0=f(r_2)$ and it is a polynomial, thus continues and differentiable. By Rolle’s Theorem $f^{\prime }(c)=0=2+3x^2+20x^4$. Since $2+3x^2+20x^4>0$ for any $x$. This contradiction proves that no two such roots exist.
17. Let $f(x)=2x-1-\sin x$. We have that $f(0)=-1<0$ and $f(\pi /2)=\pi -2>0$. Hence by the Intermediate Value Theorem $f$ has a zero in the interval $(0,\pi /2)$. Assume $f$ has two zeros $r_1$, $r_2$. Hence $f(r_1)=0=f(r_2)$. Since the function is continues and differentiable by Rolle’s theorem there $f^{\prime }(c)=0$ for some $c\in (r_1,r_2)$. Hence $0=2-\cos c$ and therefore $\cos c=2$ this contradiction shows that $f$ cannot have two roots.
18. If ${\displaystyle f}$ has two roots ${\displaystyle a<b\text{,}}$ then ${\displaystyle f\left(a\right)=f\left(b\right)=0\text{.}}$ Since ${\displaystyle f}$ is polynomial, it is continuous on ${\displaystyle \left[a,b\right]}$ and differentiable on ${\displaystyle \left(a,b\right)\text{.}}$ Rolle’s theorem implies that ${\displaystyle f^{\prime }\left(r\right)=0}$ for some ${\displaystyle a<r<b\text{.}}$ But ${\displaystyle f^{\prime }\left(r\right)=3r^2-15}$ and ${\displaystyle \left|r\right|<2\text{,}}$ so ${\displaystyle r^2<4\text{.}}$ It follows that ${\displaystyle 3r^2-15<3\left(4\right)-15=-3<0\text{.}}$ This contradiction indicates that ${\displaystyle f}$ cannot have two roots in ${\displaystyle \left[-2,2\right]\text{.}}$

20. 1. The function $f(x)=3x^2+2x+5$ is a polynomial, hence continuous and differentiable on $[-1,1]$. By the Mean Value Theorem, there is $-1<c<1$ so that

       $f(1)-f(-1)=f^{\prime }(c)1-(-1)$   and   $4=(6c+2)(-2)$

       Thus $c=-0$.

    2. The function $x)=\sqrt[3]{x}$ is continuous and differentiable for all $x$. By the Mean Value Theorem, there is $0<c,1$ so that

       ${\displaystyle f(1)-f(0)=f^{\prime }(c)(1)}$   and   ${\displaystyle 1=\frac{1}{3c^{2/3}}}$

       Thus ${\displaystyle c=\frac{1}{3\sqrt{3}}}$

21. ${\displaystyle f\left(2\right)-f\left(0\right)=3-\left(-1\right)=4\text{,}}$ ${\displaystyle f^{\prime }\left(x\right)=-\frac{2}{{\left(x-1\right)}^2}\text{.}}$

    Since ${\displaystyle f^{\prime }\left(x\right)<0}$ for all ${\displaystyle x}$ (except $x=1$), ${\displaystyle f^{\prime }\left(c\right)\left(2-0\right)}$ is always ${\displaystyle <0\text{,}}$ and hence cannot equal 4. This conclusion does not contradicts the Mean Value Theorem since ${\displaystyle f}$ is not continuous at ${\displaystyle x=1\text{.}}$

23. ${\displaystyle f\left(4\right)-f\left(1\right)=f^{\prime }\left(c\right)\left(4-1\right)}$ for some ${\displaystyle 1<c<4\text{.}}$

    But for every ${\displaystyle c}$ in ${\displaystyle \left(1,4\right)\text{,}}$ we have ${\displaystyle f^{\prime }\left(c\right)\ge 2\text{.}}$ Putting ${\displaystyle f^{\prime }\left(c\right)\ge 2}$ into the above equation and substituting ${\displaystyle f\left(1\right)=10\text{,}}$ we get ${\displaystyle f\left(4\right)=f\left(1\right)+f^{\prime }\left(c\right)\left(4-1\right)=10+3f^{\prime }\left(c\right)\ge 10+3\left(2\right)=16\text{.}}$ So the smallest possible value of ${\displaystyle f\left(4\right)}$ is 16.

24. If the function is continuous and differentiable on $[0,2]$, then by the Mean Value Theorem

    ${\displaystyle \frac{f(2)-f(0)}{2}=f^{\prime }(c)}$   for some   ${\displaystyle x\in [0,2]}$

    Hence ${\displaystyle \frac{5}{2}=f^{\prime }(x)\le 2}$ this contradiction indicates that if the function exists it must be noncontinuous or no differentiable.

25. To prove the second statement we have to show that ${\displaystyle f(a)>f(b)}$ for any ${\displaystyle a<b}$ in the interval ${\displaystyle J}$.

    We consider the interval ${\displaystyle [a,b]}$ in ${\displaystyle J}$. Since ${\displaystyle f^{\prime }(x)<0}$ for any ${\displaystyle x\in [a,b]}$, the function ${\displaystyle f}$ is differentiable in ${\displaystyle J}$, and therefore it is continuous on ${\displaystyle [a,b]}$ and differentiable on ${\displaystyle (a,b)}$. By the Mean Value Theorem

    ${\displaystyle \frac{f(b)-f(a)}{b-a}=f^{\prime }(c)}$   for some   ${\displaystyle a<c<b}$

    for this ${\displaystyle c}$, we know that ${\displaystyle f^{\prime }(c)<0}$; hence,

    ${\displaystyle \frac{f(b)-f(a)}{b-a}<0}$

    We also know that ${\displaystyle b-a>0}$, hence ${\displaystyle f(b)-f(a)<0}$ and we conclude that ${\displaystyle f(a)>f(b)}$.

26. In Definition 5.11, we indicated that a function ${\displaystyle f}$ is concave down on an interval ${\displaystyle J}$ if the graph of ${\displaystyle f}$ is above the segment from ${\displaystyle (a,f(a))}$ to ${\displaystyle (b,f(b))}$ for any ${\displaystyle a<b}$ in ${\displaystyle J}$. This is the same as saying that for any ${\displaystyle x}$ in the interval ${\displaystyle [a,b]}$ we have that ${\displaystyle f(x)}$ is below the line segment from ${\displaystyle (a,f(a))}$ to ${\displaystyle (b,f(b))}$.

    The line segment from ${\displaystyle (a,f(a))}$ to ${\displaystyle (b,f(b))}$ is given by the equation

    ${\displaystyle y=\frac{f(b)-f(a)}{b-a}(x-a)+f(b)}$   for   ${\displaystyle a\le x\le b}$.

    Thus, a function is concave down if

    ${\displaystyle f(x)>\frac{f(b)-f(a)}{b-a}(x-a)+f(b)}$

    for any ${\displaystyle a<x<b}$.

    This is equivalent to the second statement of Definition 6.7.

27. Let ${\displaystyle h(x)=-f(x)}$. If ${\displaystyle f^{″}(x)<0}$ for all ${\displaystyle x}$ in ${\displaystyle J}$, then ${\displaystyle h^{″}(x)=-f^{″}(x)>0}$ for any ${\displaystyle x}$ in ${\displaystyle J}$. Hence by part a of Theorem 6.9 we conclude that ${\displaystyle h}$ is concave up. So by Definition 6.7 part a for any ${\displaystyle a<x<b}$ in ${\displaystyle J}$

    ${\displaystyle \frac{h(x)-h(a)}{x-a}<\frac{h(b)-h(a)}{b-a}}$

    thus

    ${\displaystyle \frac{-f(x)+f(a)}{x-a}<\frac{-f(b)+f(a)}{b-a}}$

    and

    ${\displaystyle \frac{f(x)-f(a)}{x-a}>\frac{f(b)-f(a)}{b-a}}$

    By Definition 6.7 the function ${\displaystyle f}$ is concave down.