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


$(\frac{1}{16},-\frac{1}{4})$ 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.
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