## ksandoval 3 years ago how do i find the critical points of f(x) = (x-2)^5 (x+3)^4?

1. freewilly922

How would you explain a critical point? A minimum/maximum? a zero?

2. ksandoval

hmm i dont know haha... i know that i need to find the derivative first though.

3. freewilly922

A critical point of a function of a single real variable, ƒ(x), is a value x0 in the domain of ƒ where either the function is not differentiable or its derivative is 0,

4. freewilly922

So it sounds like you need to take the derivative, set it to zero and then solve for x.

5. ksandoval

and i got \[f'(x) = 5(x-2)^4(x+3)^4 + 4(x+3)^3(x-2)^5\]

6. ksandoval

embarrassing, but i honestly have no idea how to factor that. ha.

7. freewilly922

Well, the idea is to find an alternative to brute force.

8. freewilly922

Ok, when things look ugly like this, substitute something easy. Like A= x-2 and B=x+3 This gives you \[f'(X) = 5A^4B^4 +4B^3A^5\] This is easy to factor and set to zero. Then once you have solved for when THIS is zero with respect to A and B then resubstitute the A=x-2 and B=x+3 to get you final answer.