# Thread: help with critical points....

1. ## help with critical points....

hi!
ok... i was given this function:
f(x) = (x+2)^3 (1-x) (3-x)
and i have some trouble getting it's critical points...
I understand i get those by finding the points in which the derivative is 0.... right?
I got the derivative....
f'(x) = 5x^4 -8x^3 -27x^2 -44x +4

so... i equaled that to zero... and i got the four points, which are:

x1 = 2.314
x2 = -2.005
x3 = -1.995
x4 = 0.086

so the question is... does that even make any sense? I checked those points on the graph, and they dont mark the minimum values or anything... am i doing something wrong??

2. Hello buddy.

Originally Posted by c.park
hi!
ok... i was given this function:
f(x) = (x+2)^3 (1-x) (3-x)
and i have some trouble getting it's critical points...
I understand i get those by finding the points in which the derivative is 0.... right?
Yep, thats how it works.

Originally Posted by c.park
I got the derivative....
f'(x) = 5x^4 -8x^3 -27x^2 -44x +4

Uhhhhh, so close.
it's more like
5x^4 + 8x^3 - 27x^2 - 44x + 4

did you noticed the +8x^3, not -8x^3

Originally Posted by c.park
so... i equaled that to zero... and i got the four points, which are:

x1 = 2.314
x2 = -2.005
x3 = -1.995
x4 = 0.086

so the question is... does that even make any sense? I checked those points on the graph, and they dont mark the minimum values or anything... am i doing something wrong??
The solution should be

x=0.0864, x = 2.314, x = -2

Alright?

Yours
Rapha

3. By the way, those are not points, they are values of x. The "critical points" are (x, y) points with y calculated for each value of x.