Calculus – Finding Horizontal Tangent Line Using Derivatives

calculusderivatives

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?

Best Answer

Simplify $f'(x)=(x-2)(2x-1)+(1)(x^2-x-11)$

  • $f'(x) = (2x^2 -5x + 2) + (x^2 - x - 11)$
  • $f'(x) = 3x^2 - 6x - 9$

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.


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