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First Derivative Test

  • Thread starter jhodzzz
  • Start date
  • #1
15
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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?
 

Answers and Replies

  • #2
164
2
first, you need to differentiate correctly. the second differentiated term needs a what by composition...?
 
  • #3
15
0
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?
 
  • #4
164
2
no...more like f(x)=g(h(x))*m(n(x))
 
  • #5
vela
Staff Emeritus
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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.
 

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