 Terrence Riddle

2022-04-10

What is the equation of the line normal to $f\left(x\right)=2{x}^{3}-{x}^{2}-x$ at x=4? legaldaj1dn

f(4)=2*64-16-4=128-20=108
${f}^{\prime }\left(x\right)=6{x}^{2}-2x-1$
f'(4)=6*16-8-1=96-9=87
$R\left(x\right)=-\frac{1}{87}x+b$ is the desired perp.
$R\left(4\right)=108⇔-\frac{4}{87}+b=108$
$b=108+\frac{4}{87}$ abangan85s0

$f\left(x\right)=2{x}^{3}-{x}^{2}-x$ and ${x}_{0}=4$.
Find the value of the function at the given point: ${y}_{0}=f\left(4\right)=108$.
The slope of the normal line at $x={x}_{0}$ is the negative reciprocal of the derivative of the function, evaluated at $x={x}_{0}:M\left({x}_{0}\right)=-\frac{1}{{f}^{\prime }\left({x}_{0}\right)}$
Find the derivative: ${f}^{\prime }\left(x\right)={\left(2{x}^{3}-{x}^{2}-x\right)}^{\prime }=6{x}^{2}-2x-1$
Hence, $M\left({x}_{0}\right)=-\frac{1}{{f}^{\prime }\left({x}_{0}\right)}=-\frac{1}{6{x}_{0}^{2}-2{x}_{0}-1}$
Next, find the slope at the given point.
$m=M\left(4\right)=-\frac{1}{87}$
Finally, the equation of the normal line is $y-{y}_{0}=m\left(x-{x}_{0}\right)$
Plugging the found values, we get that $y-108=-\frac{x-4}{87}$
Or, more simply: $y=\frac{9400}{87}-\frac{x}{87}$.

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