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Numerade Educator



Problem 57 Hard Difficulty

Let $ f(x) = \sqrt[3]{x} $.
(a) If $ a \neq 0 $, use Equation 2.7.5 to find $ f'(a) $.
(b) Show that $ f'(0) $ does not exist.
(c) Show that $ y = \sqrt[3]{x} $ has a vertical tangent line at $ (0, 0) $. (Recall the shape of the graph of
$ f $. See Figure 1.2.13)


a) $f^{\prime}(a)=\frac{1}{3 a^{2 / 3}}=\frac{1}{3} a^{-2 / 3}$
b) $\lim _{h \rightarrow 0} \frac{1}{h^{2 / 3}}$ , DNE
(c) $\lim _{x \rightarrow 0}\left|f^{\prime}(x)\right|=\lim _{x \rightarrow 0} \frac{1}{3 x^{2 / 3}}=\infty$ and $f$ is continuous at $x=0$ (root function), so $f$ has a vertical tangent at $x=0$


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Meshal A.

November 18, 2020

The graph of f is given. State the numbers at which f is not differentiable.

Video Transcript

for this question we need to find the derivative of this function here. Um And I need to prove that it does not exist at X equals zero. And also that it has a vertical tangent line At the zero which is the origen So to find The prime effects which is the derivative. Well to use the formula the power addiction formula here where the derivative effects to the power and is and times x to the power and -1. So here our ffx when we rewrite it it would be X to the power one third and so D by D X. Ffx which is also have prime effects, is equal to one third. X rays to 1 3rd -1 which is nec X one third times x rays to negative two thirds or one third times tu brute of X squared for part B. When we plug in X equals zero for f prime we get one third time's cube root of zero squared and so three times zero is zero. And so we get 1/0 which is not divide for part C. We need to prove that there exists a vertical tangent at point 00. So at this point Her X is equal to zero. And so a prime of zero. As we proved earlier was 1/0 which means it has an infinite slope. And what else has an infinite slope. It's a vertical line. Yeah. Therefore we can conclude That have had 000 or wherever X is equal to zero. There has to be a vertical tangent. Yeah