How would I solve this problem?
Find the point where the tangent line is horizontal in the following function:
I computed the derivative: $\quad f'(x)=(x-2)(2x-1)+(1)(x^2-x-11)$.
But what would I do next?
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And find where $f'(x) = 0$
$3x^2 - 6x - 9 = 0 \iff x^2 - 2x - 3 = (x + 1)(x - 3) = 0$
That's where slope is 0, hence any line tangent at that point will be horizontal: when $x = 3$ or when $x = -1$.
So the roots (x values) of the points you need are
$x_1 = 3$, and
$x_2 = -1$.
Then find the corresponding $y$ value in the ORIGINAL equation: $$\;f(x)=(x-2)(x^2-x-11),\;$$ to determine the two points: $(x_1, y_1), (x_2, y_2)\;$ where the line tangent to $f(x)$ is horizontal.
$y_1 = f(3) = 1 \cdot -5 = -5$
$y_2 = f(-1) = (-3)(-9) = 27$
So your points are $(3, -5)$ and $(-1, 27)$.
I've included a graph of the function $\;f(x)=(x-2)(x^2-x-11)\;$(in blue), along with the two horizontal lines tangent to the function: $\;y = -5,\;\; y = 27$ (violet and "brown", respectively) to see in a "picture" what is happening here.
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:$$f'(x)=(2x^2-5x+2)+(x^2-x-11)$$
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 $(-1, 27)$