$f(x) = x|x^2-6|- \frac{3}{2} x^2 +2$
With this function I know that I need to differentiate it in order to find the local maxima and minima points. However it always confuses me a bit on what I should do when there's a modulus/absolute value within the equation. Once I've found the derivative I'm pretty certain on what to do.
Anyways what I'm thinking right now is to have three cases. Firstly when $x^2$ is greater than 6, secondly when $x^2$ is less than 6 and finally when it's equal to 6. Is this the right approach? I'm also going to test $f(-2)$ & $f(4)$ since they're the limits.
Best Answer
hint
Observe that $$-\sqrt{6}<-2<\sqrt{6}<4$$ and $$-\sqrt{6}<x\le \sqrt{6}\implies$$ $$ |x^2-6|=-x^2+6$$
$$\sqrt{6}\le x\le 4\implies |x^2-6|=x^2-6$$
Thus , at $ [\sqrt{6},4]$, $$f'(x)=3x^2-6-3=3(x^2-3)>0$$ and at $ [-2,\sqrt{6}] $, $$f'(x)=6-3x^2-3=3(1-x^2)$$
You have to check $ f(\pm 1) $.