TITLE: Calculus question taking derivative to find horizontal tangent line
QUESTION [8 upvotes]: How would I solve this problem?

Find the point where the tangent line is horizontal in the following function:
$$f(x)=(x-2)(x^2-x-11)$$

I computed the derivative: $\quad f'(x)=(x-2)(2x-1)+(1)(x^2-x-11)$.
But what would I do next?

REPLY [2 votes]: You are on your way there. Consider your first derivative $$f'(x)=(x-2)(2x-1)+(1)(x^2-x-11)$$
Let's simplify this:$$f'(x)=(2x^2-5x+2)+(x^2-x-11)$$
$$f'(x)=3x^2-6x-9$$
Now, for a horizontal line, $f'(x) = 0$. So let's solve $$3x^2-6x-9 = 0$$ $$x^2-2x-3 = 0$$
$$(x-3)(x+1) = 0$$ $x = 3 $ or $x = -1$

Hence, you answer that the tangent is horizontal at 2 points, $(3, -5) $ and $(-1, 27)$