# Max & Min

1. Jan 7, 2009

### fazal

1. The problem statement, all variables and given/known data

let f(x)=(x+1)^2(x-2)
a)Find the max and Min values of f(x)
b)Find the inflection point of concavity

2. Relevant equations

using defferentiation

3. The attempt at a solution
for part a) differentiate (x+1)^2(x-2) we get ans=3(x-1)(x+1) Plse check for me?
than f'(x)=0 therefore the points i got is x=1 and x=-1 respective??

the second part not sure..plse assist.

2. Jan 7, 2009

### jeffreydk

You are right about the local max and min. For the second part--all an inflection point is, is where the second derivative (that which we use to determine concavity) is 0. So you would have

$$f''(x)=2(x-2)+4(x+1)=0$$

Can you get it from there?

3. Jan 8, 2009

### HallsofIvy

Staff Emeritus
Make sure you answer the question asked! Max and Min occur at x= -1 and x= 1 but you haven't yet said what the max and min are.

Also what jeffreydk said about the inflection point is slightly misleading. An inflection point is a point where the first derivative changes sign. That means the the second derivative must be 0 there but that is not sufficient. You need to check that the first derivative really does change sign. For example if f(x)= x4, f'= 4x3 and f"= 12x2. f"(0)= 0 but f' does NOT change sign there so (0,0) is NOT an inflection point of f(x)= x4.