# Thread: Critical Points/Rel. Extremes HELP

1. ## Critical Points/Rel. Extremes HELP

Let f(x)=1/5x^5 + 3/2x^4 - 23x^3 + 23x^2 + 120x -56
Find :
(a) the critical points of f (if any),
(b) the open intervals on which the function is increasing or decreasing, and
(c) the location of all relative extremes (if any).

I'm kind of familiar with doing these but I have no idea on this one. The polynomial is huge. Ah!

2. Originally Posted by howyouda
Let f(x)=1/5x^5 + 3/2x^4 - 23x^3 + 23x^2 + 120x -56
Find :
(a) the critical points of f (if any),
(b) the open intervals on which the function is increasing or decreasing, and
(c) the location of all relative extremes (if any).

I'm kind of familiar with doing these but I have no idea on this one. The polynomial is huge. Ah!
I presume the problem is in finding the critical points, so I'll do that.
f(x) = 1/5x^5 + 3/2x^4 - 23x^3 + 23x^2 + 120x -56

f'(x) = x^4 + 6x^3 - 69x^2 + 46x + 120

Set this equal to 0:
x^4 + 6x^3 - 69x^2 + 46x + 120 = 0

According to the rational root theorem the possible rational roots are:
x = (+/-)1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120

This is a rather long list, but we can quickly find that
x = 2, 5
are both solutions.

We may now either continue plugging the list into the polynomial to find the other two zeros, or we may divide the polynomial by (x - 2)(x - 5) to get:
x^4 + 6x^3 - 69x^2 + 46x + 120 = (x - 2)(x - 5)(x^2 + 13x +12) = 0

So we need to solve x^2 + 13x + 12 = (x + 12)(x + 1) = 0
Thus the other two solutions are
x = -12, -1.

So the critical points of f(x) = 1/5x^5 + 3/2x^4 - 23x^3 + 23x^2 + 120x -56 are at x = -12, -1, 2, 5

-Dan