Thread: critical points of maxima

Critical points of maxima/ minima
How can i determine the critical points of this function?
f(x) = x^5 - 5x^3 - 20x - 2
And thanks ahead

Originally Posted by ^_^Engineer_Adam^_^

Critical points of maxima/ minima
How can i determine the critical points of this function?
f(x) = x^5 - 5x^3 - 20x - 2
And thanks ahead

You will find critical points where f'(x) = 0. So:
f'(x) = 5x^4 - 15x^2 - 20 = 0

This is a "bi-quadratic" equation. To make things look simpler, let y = x^2. Then we need to solve:
5y^2 - 15y - 20 = 0

5(y^2 - 3y - 4) = 0
5(y - 4)(y + 1) = 0

So y = 4 and y = -1.

Now, y = x^2 so we have x^2 = 4 and x^2 = -1.
x^2 = 4
x = -2 and x = 2.

x^2 = -1
has no real solutions.

So the critical points are at x = -2, 2.

-Dan