Minimum/Maximum: Harder functions

f(x) = (x^2 - 1)e^x

I used chain rule and I get the first derivative. Then set that to zero, but with such a complicated function how do I find the exact values of x for these critical points? Before it was simple critical points where 2x - 1 = 0 or 3x^2 = 0

but this is an exam question

f'(x)= (x^2 - 1)e^x + e^x.2x = 0

Is it just me or do I suck at algebra like this?

Is it just me or do I suck at algebra like this?

$(x^2-1)e^x + 2 x e^x = 0 \implies e^x (x^2-1+2x) = 0$

As $e^x \not= 0$, hence we can divide both sides by $e^x$

This gives, $x^2+2x-1=0$, now solve the quadratic to get the coordinates.



It is just you.

What about a function that uses trigonometry, when using the chain rule of course you will likely end up with various cosin, sine and - sine and -cosine functions, how would I set such a derivative to 0?

Solve it in the way you solve trig equations.

It will be better if you give a specific example in which you find difficulty.