Mathematics 265 Introduction to Calculus I

Solutions to Sample Midterm Exam

Sample Midterm Examination 1

Time: 3 hours
Passing grade: 55%
Total points: 64

Solutions and Marking Scheme

1. Give the exact value of ${\displaystyle cos\left(-\frac{\pi }{12}\right)\text{.}}$  (5 points)

   Solution

   ${\displaystyle \begin{array}{lll}\text{cos}\left(-\frac{\pi }{12}\right)\hfill & =\text{cos}\left(\frac{\pi }{4}-\frac{\pi }{3}\right)\hfill & \left(2\text{ pts}\right)\hfill \\ \hfill & =\text{cos}\left(\frac{\pi }{4}\right)\text{ cos}\left(\frac{\pi }{3}\right)+\text{sin}\left(\frac{\pi }{4}\right)\text{ sin}\left(\frac{\pi }{3}\right)\hfill & \left(2\text{ pts}\right)\hfill \\ \hfill & =\frac{1}{2\sqrt{2}}+\frac{\sqrt{3}}{2\sqrt{2}}=\frac{1+\sqrt{3}}{2\sqrt{2}}\hfill & \left(1\text{ pt}\right)\hfill \end{array}}$

2. Let ${\displaystyle f\left(x\right)=2x^2-5}$ and ${\displaystyle g\left(x\right)=\frac{x+5}{2x-9}}$.  (6 points)

   Find the composite functions and their corresponding domains.

   Solution

   1.  

      ${\displaystyle \begin{array}{ll}f(g(x))\hfill & =2{(g(x))}^2-5\hfill \\ \hfill & =2{\left(\frac{x+5}{2x-9}\right)}^2-5\hfill \\ \hfill & =\frac{2{(x+5)}^2}{{(2x-9)}^2}-\frac{5{(2x-9)}^2}{{(2x-9)}^2}\hfill \\ \hfill & =\frac{2{(x+5)}^2-5{(2x-9)}^2}{{(2x-9)}^2}\hfill \\ \hfill & =\frac{2(x^2+10x+25)-5(4x^2-36x+81)}{{(2x-9)}^2}\hfill \\ \hfill & =\frac{-18x^2+200x-355}{{(2x-9)}^2}\hfill \end{array}\left(2\text{ pts}\right)}$

      For the domain: ${\displaystyle 2x-9\ne 0\text{;}x\ne \frac{9}{2}\text{.}}$

      Thus all real numbers except ${\displaystyle \frac{9}{2}}$ in set notation ${\displaystyle ℝ-\left\{\frac{9}{2}\right\}\left(1\text{ pt}\right)}$

   2.  

      ${\displaystyle \begin{array}{ll}g(f(x))\hfill & =\frac{f(x)+5}{2f(x)-9}\hfill \\ \hfill & =\frac{2x^2-5+5}{2(2x^2-5)-9}\hfill \\ \hfill & =\frac{2x^2}{4x^2-19}\hfill \end{array}\left(2\text{ pts}\right)}$

      For the domain: ${\displaystyle 4x^2-19\ne 0\text{;}{x}^2\ne \frac{19}{4}\text{;}x\ne \pm {\left(\frac{19}{4}\right)}^{\frac{1}{2}}}$

      Thus all real numbers except ${\displaystyle {\left(\pm \frac{19}{4}\right)}^{\frac{1}{2}}}$ in set notation

      ${\displaystyle ℝ-\left\{\pm {\left(\frac{19}{4}\right)}^{\frac{1}{2}}\right\}\left(1\text{ pt}\right)}$

3. Give a labeled graph of the function ${\displaystyle g\left(x\right)=3{\left(x+4\right)}^2}$ by starting with the graph of a basic function, and then applying the appropriate transformations.  (5 points)

   Explain the procedure you are using.

   Note: No credit will be given if any other method is used.

   Solution

   Basic function: ${\displaystyle u\left(x\right)=x^2\left(1\text{ pt}\right)}$

   Shift to the left by 4 units: ${\displaystyle u\left(x+4\right)={\left(x+4\right)}^2\left(1\text{ pt}\right)}$

   Vertical Stretch by 3 units: ${\displaystyle g\left(x\right)=3{\left(x+4\right)}^2\left(1\text{ pt}\right)}$

   Graph: ${\displaystyle \left(2\text{ pts}\right)}$

   

4. Evaluate each of the limits below. If a limit does not exist explain why.  (16 points)
   1. ${\displaystyle \underset{x\to 2}{lim}\left(3x^2-2x+1\right)}$
   2. ${\displaystyle \underset{x\to -2}{lim}\frac{3x^2-2x-16}{{\left(x+2\right)}^2}}$
   3. ${\displaystyle \underset{x\to 3}{lim}\frac{2x^2-x+1}{x-3}}$
   4. ${\displaystyle \underset{x\to 2}{lim}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}}$
   5. ${\displaystyle \underset{x\to 0}{lim}\frac{sin\left(3x\right)}{x^2-x}}$

   Solution

   1. The polynomial ${\displaystyle 3x^2-2x+1}$ is continuous at 2. ${\displaystyle \left(1\text{ pt}\right)}$

      Thus

      ${\displaystyle \underset{x\to 2}{lim}\left(3x^2-2x+1\right)=3{\left(2\right)}^2-2\left(2\right)+1=9\left(2\text{ pts}\right)}$

   2. The limit

      ${\displaystyle \underset{x\to -2}{lim}\frac{3x^2-2x-16}{{\left(x+2\right)}^2}=\underset{x\to -2}{lim}\frac{\left(3x-8\right)\left(x+2\right)}{{\left(x+2\right)}^2}=\underset{x\to -2}{lim}\frac{3x-8}{x+2}}$

      does not exist, ${\displaystyle \left(1\text{ pt}\right)}$

      because

      ${\displaystyle \underset{x\to -2^{+}}{lim}\frac{3x-8}{x+2}=-\infty }$

      since

      ${\displaystyle \underset{x\to -2}{lim}3x-8=3\left(-2\right)-8=-14<0}$

      and

      ${\displaystyle \underset{x\to -2^{+}}{lim}\frac{1}{x+2}=\infty \left(1\text{ pt}\right)}$

      this last limit is infinity because

      ${\displaystyle x+2>0\text{for}x\to -2^{+}\text{and}\underset{x\to -2^{+}}{lim}x+2=0}$

      Similarly

      ${\displaystyle \underset{x\to -2^{-}}{lim}\frac{3x-8}{x+2}=\infty }$

      since

      ${\displaystyle \underset{x\to -}{lim}3x-8=3\left(-2\right)-8=-14<0}$

      and

      ${\displaystyle \underset{x\to -2^{-}}{lim}\frac{1}{x+2}=-\infty \left(1\text{ pt}\right)}$

      because

      ${\displaystyle x+2<0\text{for}x\to -2^{-}\text{and}\underset{x\to -2^{-}}{lim}x+2=0}$

   3. The limit

      ${\displaystyle \underset{x\to 3}{lim}\frac{2x^2-x+1}{x-3}}$

      does not exist, ${\displaystyle \left(1\text{ pt}\right)}$

      because we have that

      ${\displaystyle \underset{x\to 3^{+}}{lim}\frac{2x^2-x+1}{x-3}=\infty }$

      since

      ${\displaystyle \underset{x\to 3^{+}}{lim}2x^2-x+1=2\left(9\right)-3+1=16>0}$

      and

      ${\displaystyle \underset{x\to 3^{+}}{lim}\frac{1}{x-3}=\infty \left(1\text{ pt}\right)}$

      this last limit is infinity because

      ${\displaystyle x-3>0\text{for}x\to 3^{+}\text{and}\underset{x\to 3^{+}}{lim}x-3=0}$

      Similarly,

      ${\displaystyle \underset{x\to 3^{-}}{lim}\frac{2x^2-x+1}{x-3}=-\infty }$

      since

      ${\displaystyle \underset{x\to 3^{+}}{lim}2x^2-x+1=2\left(9\right)-3+1=16>0}$

      and

      ${\displaystyle \underset{x\to 3^{-}}{lim}\frac{1}{x-3}=-\infty \left(1\text{ pt}\right)}$

      because

      ${\displaystyle x-3<0\text{for}x\to 3^{-}\text{and}\underset{x\to 3^{-}}{lim}x-3=0}$

   4. We rationalize the numerator and denominator and obtain

      ${\displaystyle \begin{array}{lll}\underset{x\to 2}{\lim }\hfill & \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}=\hfill & \hfill \\ \hfill & \underset{x\to 2}{\lim }\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}\cdot \frac{\left(\sqrt{6-x}+2\right)\left(\sqrt{3-x}+1\right)}{\left(\sqrt{6-x}+2\right)\left(\sqrt{3-x}+1\right)}\hfill & (1\text{ pt)}\hfill \\ \hfill & =\underset{x\to 2}{\lim }\frac{\left(2-x\right)\left(\sqrt{3-x}+1\right)}{\left(2-x\right)\left(\sqrt{6-x}+2\right)}\hfill & (1\text{ pt)}\hfill \\ \hfill & =\underset{x\to 2}{\lim }\frac{\sqrt{3-x}+1}{\sqrt{6-x}+2}=\frac{2}{4}=\frac{1}{2}\hfill & (1\text{ pt)}\hfill \end{array}}$

   5. ${\displaystyle \begin{array}{llllllll}\hfill \underset{x\to 0}{lim}\frac{sin\left(3x\right)}{x^2-x}& =\underset{x\to 0}{lim}\frac{3sin\left(3x\right)}{3x}\frac{1}{x-1}& \hfill & \left(2\text{ pts}\right)& \hfill & & \hfill & \\ \hfill & =3\left(1\right)\left(-1\right)=-3& \hfill & \left(2\text{ pts}\right)& \hfill & & \hfill & \end{array}}$
5. Compute the derivatives of each of the functions below. You may not need to simplify your answers.  (8 points)
   1. ${\displaystyle y=4x^5+3x^4-6x^3+6}$
   2. ${\displaystyle y=\frac{2x-16}{{\left(x+3\right)}^2}}$
   3. ${\displaystyle y=sin\left(2x^2-x+1\right)}$ ${\displaystyle y={\left(-4x^3-x^2+3x+7\right)}^4}$

   Solution

   1. By the Power Rule,

      ${\displaystyle \frac{d}{dx}4x^5+3x^4-6x^3+6=20x^4+12x^3-18x^2\left(2\text{ pts}\right)}$

   2. By the Quotient Rule,

      ${\displaystyle \begin{array}{lll}\frac{d}{dx}\frac{2x-16}{{\left(x+3\right)}^2}\hfill & =\frac{2{\left(x+3\right)}^2-\left(2x-16\right)2\left(x+3\right)}{{\left(x+3\right)}^4}\hfill & \hfill \\ \hfill & =\frac{2\left(19-x\right)}{{\left(x+3\right)}^3}\hfill & \left(2\text{ pts}\right)\hfill \end{array}}$

   3. By the Chain and Power rules,

      ${\displaystyle \begin{array}{l}\frac{d}{dx}\text{ sin}\left(2x^2-x+1\right)=\hfill \\ \cos \left(2x^2-x+1\right)\left(4x-1\right)\left(2\text{ pts}\right)\hfill \end{array}}$

   4. By the General Power Rule,

      ${\displaystyle \begin{array}{l}\frac{d}{dx}{\left(-4x^3-x^2+3x+7\right)}^4=\\ 4{\left(-4x^3-x^2+3x+7\right)}^3\left(-12x^2-2x+3\right)\end{array}}$

6. Find the values of ${\displaystyle x}$ such that tangent line to the curve ${\displaystyle f\left(x\right)=3x^2+4x-3}$ is perpendicular to the line ${\displaystyle y=6x+2\text{.}}$  (4 points)

   The slope of the line tangent to ${\displaystyle f\left(x\right)=3x^2+4x-3}$ is ${\displaystyle f^{\prime }\left(x\right)=6x+4\text{.}\left(1\text{ pt}\right)}$

   The slope of the line ${\displaystyle y=6x+2}$ is ${\displaystyle 6\text{.}\left(1\text{ pt}\right)}$

   So we want ${\displaystyle x}$ such that ${\displaystyle 6x+4=-\frac{1}{6}\text{,}\left(1\text{ pt}\right)}$

   solving for ${\displaystyle x}$, we find

   ${\displaystyle x=-\frac{25}{36}\left(1\text{ pt}\right)}$

7. Gravel is being dumped from a conveyor belt at a rate of ${\displaystyle 0\text{.}5}$ m ${\displaystyle {}^3}$ / min. It forms a pile in shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is ${\displaystyle 4}$ m high?   (5 points)

   Solution

   Volume of sand: ${\displaystyle V=\frac{\pi r^{2}h}{3}\left(1\text{ pt}\right)}$

   Since ${\displaystyle h=}$ diameter ${\displaystyle =2r}$, the volume is ${\displaystyle V=\frac{\pi h^3}{12}\left(1\text{ pt}\right)}$

   differentiating, we find ${\displaystyle \frac{dV}{dt}=\frac{\pi }{12}\left(3h^2\right)\frac{dh}{dt}=\frac{\pi h^2}{4}\frac{dh}{dt}\left(1\text{ pt}\right)}$

   Hence, for ${\displaystyle h=4}$, solving for ${\displaystyle \frac{dh}{dt}}$, we have

   ${\displaystyle \frac{dh}{dt}=\frac{4}{\pi {\left(4\right)}^2}\left(\frac{1}{2}\right)=\frac{1}{8\pi }}$ m/min ${\displaystyle \left(2\text{ pts}\right)}$

8. Find ${\displaystyle y^{\prime }}$ using implicit differentiation:  (5 points)

   ${\displaystyle x^{3}y+x{y}^2=4xy+7\text{.}}$

   Solution

   ${\displaystyle 3x^{2}y+x^3{y}^{\prime }+y^2+2xy{y}^{\prime }=4\left(y+x{y}^{\prime }\right)\left(3\text{ pts}\right)}$

   Solving for ${\displaystyle y^{\prime }}$:

   ${\displaystyle y^{\prime }=\frac{4y-3x^{2}y-y^2}{x^3+2xy-4x}\left(2\text{ pts}\right)}$

9. Use differentials to find the approximate value of ${\displaystyle \sqrt{9\text{.}2}\text{.}}$  (5 points)

   Solution

   ${\displaystyle \sqrt{9\text{.}2}=\sqrt{9+0\text{.}2}\left(1\text{ pt}\right)}$

   Let ${\displaystyle f\left(x\right)=\sqrt{x}}$, and ${\displaystyle \Delta x=0\text{.}2\text{.}\left(1\text{ pt}\right)}$

   Hence,

   ${\displaystyle \sqrt{9\text{.}2}\approx f\left(3\right)+f^{\prime }\left(3\right)\text{Δ}x\left(1\text{ pt}\right)}$

   ${\displaystyle =\sqrt{9}+\frac{0\text{.}2}{2\sqrt{9}}=3+\frac{1}{30}\left(2\text{ pts}\right)}$

10. Sketch the graph of a single function ${\displaystyle f\left(x\right)}$ that satisfies all of the conditions listed below.  (5 points)
    • the limit ${\displaystyle \underset{x\to 0}{lim}f\left(x\right)}$ does not exist.
    • ${\displaystyle f\left(0\right)=0\text{.}}$
    • ${\displaystyle \underset{x\to \infty }{lim}f\left(x\right)=-1\text{.}}$
    • the function ${\displaystyle f}$ is not differentiable at ${\displaystyle x=-2\text{.}}$

    Answers will vary. Award yourself one point for each condition, and one point for sketching the graph of an actual function. For example, the graph below would be given 5 points:

    

    Sketches with overlapping lines, such as the one below, are not graphs of functions, since the vertical line test fails in this case.