First Derivative Test for f(x) = (1-x)^2(1+x)^3: Extrema, Intervals, and Values

[jhodzzz]

Homework Statement

I have to get the following:
- relative extrema of f
- values of f at which the relative extrema occurs
- intervals on which f is increasing
- intervals on which f is decreasing

when f(x) = (1-x)² (1+x)³

Homework Equations

Now when get to have the first derivative by multiplication rule f'(x) = g(x)*h'(x)+h(x)*g'(x):
f'(x) = ((1-x)²)(3(1+x)²)+((1+x)³)(2(1-x))
is it correct to say that f'(x)=0 when x=1 or x=-1?

and if it is, by substituting 1 and -1 to f(x), i'll arrive on ordered pairs' (1,0),(-1,0) which are on a vertical line. when i checked if the interval -1 < x < 1 is increasing or decreasing, i arrived at an answer that it is increasing which is not possible considering the locations of the two critical points.

Where did I go wrong?

 

[xaos]

first, you need to differentiate correctly. the second differentiated term needs a what by composition...?

 

[jhodzzz]

you mean this differentiation: f'(x)=((1-x)²)(3(1+x)²)+((1+x)³)(2(1-x))

I arrived at that considering f(x)=g(x)*h(x) such that g(x)=(1-x)² and h(x)=(1+x)³

so applying the multiplication rule, i should have that answer.
Do I still need to simplify it further?? will the factors vary by then?

 

[xaos]

no...more like f(x)=g(h(x))*m(n(x))

 

[vela]

You need to apply the chain rule to get the second term correct.

You are correct that the function has zeros as x=1 and x=-1, but there's another zero that you won't see until you simplify the expression.