Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 5

Applications of the Derivative

Learning from Mistakes

There are mistakes in each of the following solutions. Identify the errors, and give the correct answer.

1. Sketch the graph of the function ${\displaystyle f\left(x\right)={\left(x-1\right)}^{3∕2}}$.

   Erroneous Solution

   ${\displaystyle f^{\prime }\left(x\right)=3/2{\left(x-1\right)}^{1∕2}}$ the critical number is ${\displaystyle x=1}$, because ${\displaystyle f^{\prime }\left(1\right)=0}$.

   Since ${\displaystyle f^{\prime }\left(x\right)>0}$ for all $x$, the function is increasing for all $x$.

   Since ${\displaystyle f^{″}\left(x\right)={\displaystyle \frac{3}{4{\left(x-1\right)}^{-1∕2}}}}$, the critical number is ${\displaystyle x=1}$, because ${\displaystyle f^{″}\left(1\right)}$ is undefined.

   Since ${\displaystyle f^{″}\left(x\right)>0}$ for all $x$, the function is concave up for all $x$.

   

   Figure 5.25. Erroneous solution to “Learning from Mistakes” Question ${\displaystyle 1}$

2. Find the local extreme values of the function ${\displaystyle g\left(x\right)=x^6+6x^4}$.

   Erroneous Solution

   ${\displaystyle g^{\prime }\left(x\right)=6x^5+4x^3=2x^3\left(x^2+4\right)=0}$ if ${\displaystyle x=0}$; therefore, ${\displaystyle x=0}$ is the only critical number.

   ${\displaystyle g^{″}\left(x\right)=30x^4+12x^2\text{and}{g}^{″}\left(0\right)=0\text{.}}$

   By the second derivative test, the function has neither a minimum nor a maximum; hence, it has an inflection point at ${\displaystyle x=0}$.

3. Find the extreme values of the function ${\displaystyle h\left(x\right)=x^3-6x}$ on the interval ${\displaystyle \left[0,2\right]}$.

   Erroneous Solution

   To find the critical points ${\displaystyle h^{\prime }\left(x\right)=3x^2-6=0}$, we solve for $x$. The critical numbers are ${\displaystyle x=\sqrt{2}}$ and ${\displaystyle x=-\sqrt{2}}$. Therefore,

   ${\displaystyle h\left(\sqrt{2}\right)=\left(\sqrt{2}\right)3-6\sqrt{2}\approx -5\text{.}656}$

   and

   ${\displaystyle h\left(-\sqrt{2}\right)=\left(-\sqrt{2}\right)3+6\sqrt{2}\approx 5\text{.}656\text{.}}$

   At the endpoints ${\displaystyle h\left(0\right)=0}$ and ${\displaystyle h\left(2\right)=-4}$.

   So the absolute minimum value is

   ${\displaystyle {\sqrt{2}}^3-{\displaystyle \frac{6}{\sqrt{2}}}\approx -5\text{.}656\text{ at }x=\sqrt{2}\text{,}}$

   and the absolute maximum value is

   ${\displaystyle -{\sqrt{2}}^3+{\displaystyle \frac{6}{\sqrt{2}}}\approx 5\text{.}656\text{ at }x=-\sqrt{2}\text{.}}$

4. Find a number greater than or equal to ${\displaystyle 2}$ such that the sum of the number and its reciprocal is as small as possible.

   Erroneous Solution

   Let $x$ be such a number. Hence, we want to minimize the function ${\displaystyle f\left(x\right)=x+{\displaystyle \frac{1}{x}}}$.

   ${\displaystyle f^{\prime }\left(x\right)=1-{\displaystyle \frac{1}{x^2}}=0\text{ if }x=1\text{ or }x=-1\text{.}}$

   Since

   ${\displaystyle f^{″}\left(x\right)={\displaystyle \frac{2}{x^3}}}$

   we have ${\displaystyle f^{″}\left(1\right)>0}$ and ${\displaystyle f^{″}\left(-1\right)<0}$. Then, by the second derivative test, $f$ has a minimum at ${\displaystyle x=1}$.