Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.3 - How Derivatives Affect the Shape of a Graph - 4.3 Exercises: 21

Answer

$(\frac{1}{16},-\frac{1}{4})$ is the local minimum of $f$. There is no local maximum.

Work Step by Step

$$f(x)=\sqrt x-\sqrt[4]x$$ $$f(x)=x^{1/2}-x^{1/4}$$ 1) Find $f'(x)$ and $f''(x)$ $$f'(x)=\frac{1}{2}x^{-1/2}-\frac{1}{4}x^{-3/4}$$ $$f'(x)=\frac{1}{2}x^{-3/4}x^{1/4}-\frac{1}{4}x^{-3/4}$$ $$f'(x)=\frac{1}{2}x^{-3/4}(x^{1/4}-\frac{1}{2})$$ $$f'(x)=\frac{\sqrt[4]x-\frac{1}{2}}{2\sqrt[4]{x^3}}$$ $$f''(x)=\frac{1}{2}\frac{-1}{2}x^{-3/2}-\frac{1}{4}\frac{-3}{4}x^{-7/4}$$ $$f''(x)=-\frac{1}{4\sqrt{x^3}}+\frac{3}{16\sqrt[4]{x^7}}$$ 2) The First Derivative Method Since there exists $\sqrt[4]x$ and $\sqrt[4]{x^3}$ in $f'$, the domain of $f'$ here is $[0,+\infty)$. In other words, we would not consider the interval $(-\infty,0)$. Also, we notice that the denominator $2\sqrt[4]{x^3}\gt0$ for all $x\in(0,+\infty)$. Now we need to build a sign table for $f'(x)$. Considering all the facts above, we only need to take one point where the change of signs happen, which is $\sqrt[4]x=\frac{1}{2}$, or $x=\frac{1}{16}$. The table has been built in the image below. As we observe the table, $f'$ only changes from negative to positive at $x=\frac{1}{16}$, so $f$ has a local minimum at $x=\frac{1}{16}$. $f'$ does not change from positive to negative anytime. $$f(\frac{1}{16})=\sqrt{\frac{1}{16}}-\sqrt[4]{\frac{1}{16}}=\frac{1}{4}-\frac{1}{2}=-\frac{1}{4}$$ Therefore, $(\frac{1}{16},-\frac{1}{4})$ is the local minimum of $f$. There is no local maximum. 3) The Second Derivative Method Find $f'(x)=0$ $$\frac{\sqrt[4]x-\frac{1}{2}}{2\sqrt[4]{x^3}}=0$$ $$\sqrt[4]x-\frac{1}{2}=0$$ $$\sqrt[4]x=\frac{1}{2}$$ $$x=\frac{1}{16}$$ Now the job is find out sign of $f''(\frac{1}{16})$. $$f''(\frac{1}{16})=-\frac{1}{4\sqrt{\frac{1}{16^3}}}+\frac{3}{16\sqrt[4]{\frac{1}{16^7}}}=-16+24=8$$ $f''(\frac{1}{16})=8\gt0$, so $f$ has a local minimum at $x=\frac{1}{16}$, which is $(\frac{1}{16},-\frac{1}{4})$. $f$ has no local maximum. 4) Comparison Both tests give out the same result. However, for me, I find the calculations in the First Test less daunting than those in the Second Derivative Test. So I still prefer the First Derivative Test.
Small 1493898382
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.