Lennie Davis

2022-01-12

Find the point where the tangent line is horizontal in the following function:

$f\left(x\right)=(x-2)({x}^{2}-x-11)$

Buck Henry

Beginner2022-01-13Added 33 answers

Simplify ${f}^{\prime}\left(x\right)=(x-2)(2x-1)+\left(1\right)({x}^{2}-x-11)$

${f}^{\prime}\left(x\right)=(2{x}^{2}-5x+2)+({x}^{2}-x-11)$

${f}^{\prime}\left(x\right)=3{x}^{2}-6x-9$

And find where${f}^{\prime}\left(x\right)=0$

$3{x}^{2}-6x-9=0\le ft\Rightarrow {x}^{2}-2x-3=(x+1)(x-3)=0$

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:

$f\left(X\right)=(x-2)({x}^{2}-x-11)$

to determine the two points:$({x}_{1},{y}_{1}),({x}_{2},{y}_{2})$ where the line tangent to f(x) is horizontal.

${y}_{1}=f\left(3\right)=1\cdot -5=-5$

${y}_{2}=f(-1)=(-3)(-9)=27$

So your points are$(3,-5)$ and $(-1,27)$

And find where

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

Then find the corresponding y value in the ORIGINAL equation:

to determine the two points:

So your points are

stomachdm

Beginner2022-01-14Added 33 answers

You are on your way there. Consider your first derivative

${f}^{\prime}\left(x\right)=(x-2)(2x-1)+\left(1\right)({x}^{2}-x-11)$

Let's simplify this:

${f}^{\prime}\left(x\right)=(2{x}^{2}-5x+2)+({x}^{2}-x-11)$

${f}^{\prime}\left(x\right)=3{x}^{2}-6x-9$

Now, for a horizontal line,${f}^{\prime}\left(x\right)=0$ . So let's solve

$3{x}^{2}-6x-9=0$

${x}^{2}-2x-3=0$

$(x-3)(x+1)=0$

$x=3$ or $x=-1$

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

Let's simplify this:

Now, for a horizontal line,

Hence, you answer that the tangent is horizontal at 2 points,

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