Question

Find the local extreme of f and the intervals on which f is increasing or is decreasing.

f(x) = x^2 [times] (x^2 - 4)^(1/3)

I'm having a heck of a time trying to get the derivative for this equation.

Answer 1

\[x^2\times(x^4-4)^{\frac{1}{3}}\]right?

Answer 2

\[x^{2} * \sqrt[3]{x^{2} - 4}\]

I understand the formula to use (product rule) but everytime I right it out it comes out incorrect.

Answer 3

Nancy, I've updated it using the Equation tab to make it more readable. I'm afraid your guess was incorrect, entirely due to my bad equation writing. Sorry!

Answer 4

\begin{eqnarray*}&&f(x) = x^2 \left(x^2 - 4\right)^{1/3} \\\Rightarrow&& f'(x) = \left(x^2\right)' \left(x^2 - 4\right)^{1/3}+x^2\left(\left(x^2 - 4\right)^{1/3}\right)' \quad\quad\quad\quad\quad\mbox{Product rule}\\\Rightarrow&&f'(x) = 2x \left(x^2 - 4\right)^{1/3}+x^2\frac{1}{3}\left(x^2 - 4\right)^{-2/3}\left(x^2-4\right)'\quad\quad\mbox{Chain rule}\\\Rightarrow&&f'(x) = 2x\left(x^2 - 4\right)^{1/3}+\frac{1}{3}\frac{2x^3}{\left(x^2-4\right)^{2/3}}\\\Rightarrow&&f'(x) = \frac{6x\left(x^2 - 4\right)^{1/3}\left(x^2 - 4\right)^{2/3}+2x^3}{3\left(x^2 - 4\right)^{2/3}}\\\Rightarrow&&f'(x) = \frac{6x(x^2 - 4)+2x^3}{3\left(x^2 - 4\right)^{2/3}} = \frac{8x\left(x^2 - 3\right)}{3\left(x^2 - 4\right)^{2/3}}.\end{eqnarray*}

Answer 5

\[(x^2*\sqrt[3]{x^2-4})^{3}=x^8*(x^2-4)\]\[x^{10}-4x^8= x^8(x^2-4)\]
\[x^2-4=0\]\[x=\pm \sqrt{4}\]\[x=\pm2\]x=0