Lennie Davis

2022-01-12

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

Buck Henry

Simplify ${f}^{\prime }\left(x\right)=\left(x-2\right)\left(2x-1\right)+\left(1\right)\left({x}^{2}-x-11\right)$
${f}^{\prime }\left(x\right)=\left(2{x}^{2}-5x+2\right)+\left({x}^{2}-x-11\right)$
${f}^{\prime }\left(x\right)=3{x}^{2}-6x-9$
And find where ${f}^{\prime }\left(x\right)=0$
$3{x}^{2}-6x-9=0\le ft⇒{x}^{2}-2x-3=\left(x+1\right)\left(x-3\right)=0$
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:
$f\left(X\right)=\left(x-2\right)\left({x}^{2}-x-11\right)$
to determine the two points: $\left({x}_{1},{y}_{1}\right),\left({x}_{2},{y}_{2}\right)$ where the line tangent to f(x) is horizontal.
${y}_{1}=f\left(3\right)=1\cdot -5=-5$
${y}_{2}=f\left(-1\right)=\left(-3\right)\left(-9\right)=27$
So your points are $\left(3,-5\right)$ and $\left(-1,27\right)$

stomachdm

${f}^{\prime }\left(x\right)=\left(x-2\right)\left(2x-1\right)+\left(1\right)\left({x}^{2}-x-11\right)$
Let's simplify this:
${f}^{\prime }\left(x\right)=\left(2{x}^{2}-5x+2\right)+\left({x}^{2}-x-11\right)$
${f}^{\prime }\left(x\right)=3{x}^{2}-6x-9$
Now, for a horizontal line, ${f}^{\prime }\left(x\right)=0$. So let's solve
$3{x}^{2}-6x-9=0$
${x}^{2}-2x-3=0$
$\left(x-3\right)\left(x+1\right)=0$
$x=3$ or $x=-1$
Hence, you answer that the tangent is horizontal at 2 points, $\left(3,-5\right)$ and $\left(-1,27\right)$

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