8
$\begingroup$

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?

$\endgroup$
13
$\begingroup$

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.


enter image description here

$\endgroup$
6
  • $\begingroup$ I see I simplified the function to $3x^2-6x-9$ then I got $3(x-3)(x+1)$ then would I set it to zero $\endgroup$ – Fernando Martinez Feb 2 '13 at 17:36
  • $\begingroup$ Exactly, solve for x, and then and we use the original equation to determine the y-value corresponding to each of the two x-value solutions. See the complete problem above. $\endgroup$ – amWhy Feb 2 '13 at 17:52
  • $\begingroup$ I have a quick question I get how you got (3,-5) by plugging 3 into the orgional function but whenever I plug in -1 I get 33.I must be doing something wrong $\endgroup$ – Fernando Martinez Feb 2 '13 at 17:53
  • $\begingroup$ Oh I see never mind. Thanks for your answer. $\endgroup$ – Fernando Martinez Feb 2 '13 at 17:54
  • 1
    $\begingroup$ You did a good job.. on a homework problem! $\endgroup$ – richard1941 Mar 30 '18 at 12:48
2
$\begingroup$

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 $(-1, 27)$

$\endgroup$
3
  • $\begingroup$ Yes it make sense then do I have to plug the x values to find the y value of the point? $\endgroup$ – Fernando Martinez Feb 2 '13 at 17:38
  • $\begingroup$ yes, as edited. $\endgroup$ – bryanblackbee Feb 2 '13 at 17:46
  • $\begingroup$ gracias for your answer. $\endgroup$ – Fernando Martinez Feb 2 '13 at 17:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.