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

Beginner2022-04-11Added 9 answers

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.

$R\left(4\right)=108\iff -\frac{4}{87}+b=108$

$b=108+\frac{4}{87}$

f'(4)=6*16-8-1=96-9=87

abangan85s0

Beginner2022-04-12Added 16 answers

$f\left(x\right)=2{x}^{3}-{x}^{2}-x$ 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: ${f}^{\prime}\left(x\right)={(2{x}^{3}-{x}^{2}-x)}^{\prime}=6{x}^{2}-2x-1$

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 $y-{y}_{0}=m(x-{x}_{0})$

Plugging the found values, we get that $y-108=-\frac{x-4}{87}$

Or, more simply: $y=\frac{9400}{87}-\frac{x}{87}$.

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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