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