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I have function defines as :

$$f(y)=\sqrt[3]{x^2-x^3}+x$$

and need to find points where the tangent line is parallel to y-axis and point where tangent line does not exist. I found points where derivation is infinity [0,1], which should be the points where tangent line is parallel to y-axis. But i have problems to determinate how to find points where tangent line does not exist. Any advices?

janerosie
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  • take the derivative of the function and see where is the derivative function undefined. –  Jan 11 '20 at 12:41
  • that is in 0 and 1, but how do i know if tangent line exist? – janerosie Jan 11 '20 at 12:43
  • if tangent line exists at a point $x_0$ then that point would be a defined in $dom(\frac{d}{dx}\left(f\left(x\right)\right))$,e.g. $x_0 ∈ D_ {f'(x)}$ in your example tangent line does exist at $\left(-∞,+∞\right)-\left{0,1\right}$ –  Jan 11 '20 at 12:45
  • excuse me but iam not really sure what does it mean – janerosie Jan 11 '20 at 12:48
  • tell me what is unclear for you –  Jan 11 '20 at 12:49
  • Well, does it mean that in every point of function there is a tangent line, except 1 and 0, and x=0 and x=1 are parallel to y-axis? – janerosie Jan 11 '20 at 12:53
  • I've added some notes in my post, look at it –  Jan 11 '20 at 13:09

1 Answers1

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$$\frac{d}{dx}\left(\sqrt[3]{x^{2}-x^{3}}+x\right)=\frac{2x-3x^{2}}{3\sqrt[3]{\left(x^{2}-x^{3}\right)^{2}}}+1$$

clearly the function is undefined for $x=0,1$

Using limit definition for derivatives we have:

$$f'_{+}\left(0\right)=\lim_{x \to 0^+}\frac{f\left(x\right)-f\left(0\right)}{x-0}$$$$=\lim_{x \to 0^+}\frac{\left(\sqrt[3]{x^{2}-x^{3}}+x\right)-0}{x-0}$$$$=\lim_{x \to 0^+}\frac{\left(\sqrt[3]{x^{2}-x^{3}}+x\right)}{x}$$$$=\lim_{x \to 0^+}\frac{x\left(\sqrt[3]{\frac{1}{x}-1}+1\right)}{x}=1+\sqrt[3]{\lim_{x \to 0^+}\frac{1}{x}-1}=+∞$$ also for the other side we have: $$f'_{-}\left(0\right)=\lim_{x \to 0^-}\frac{f\left(x\right)-f\left(0\right)}{x-0}$$$$=\lim_{x \to 0^-}\frac{\left(\sqrt[3]{x^{2}-x^{3}}+x\right)-0}{x-0}$$$$=\lim_{x \to 0^-}\frac{\left(\sqrt[3]{x^{2}-x^{3}}+x\right)}{x}$$$$=\lim_{x \to 0^-}\frac{x\left(\sqrt[3]{\frac{1}{x}-1}+1\right)}{x}=1+\sqrt[3]{\lim_{x \to 0^-}\frac{1}{x}-1}=-∞$$

do the same for $x=1$ and you will see that the tangent line at these two points is parallel to vertical axis.