Find and classify the critical point(s)
f(x) = x^3 - x^2 - x

I know I need to take the derivative which would give me
3x^2 - 2x - 1 = 0
and I think I need to set them to zero to find the points but I'm not sure? would it be like
3x^2 = 0 and 2x-1 = 0 and then solve?

Question

Find and classify the critical point(s)
f(x) = x^3 - x^2 - x

I know I need to take the derivative which would give me
3x^2 - 2x - 1 = 0
and I think I need to set them to zero to find the points but I'm not sure? would it be like
3x^2 = 0 and 2x-1 = 0 and then solve?

samgrace · Answer

Critical points are where the graph intersects the dependent axis, the x axis. Therefore derivatives are not the correct course of action, these would give you the minimum/maximum points.

samgrace · Answer

Put the graph equal to zero and solve for x by factoring out x

samgrace · Answer

$$f(x) = x^3 - x^2 - x => 0 = x(x^2 - x - 1) $$

$$x = (-b \pm \sqrt(b^2-4ac))/2a$$

anonymous · Answer

wait, what numbers do I plug in for a, b, c then? 3, 2, 1?

anonymous · Answer

Take the derivative as you did and then factor the derivative. 3x^2 - 2x - 1 = 0 is just a regular quadratic that can be factored. If you factor it correctly, the values for x that you find that makes the derivative equal 0 will be the x-coordinates of the critical points youre looking for.

anonymous · Answer

i think the critical points are where the derivative is equal to zero, in the case of a polynomial
take the derivative, set it equal to zero and solve for $x$

anonymous · Answer

if you cannot see how to factor it, or if it does not factor, use the quadratic formula
i suck at factoring, so i almost always use it