zdebe5l8

2022-04-08

What is the equation of the line normal to $f\left(x\right)=\frac{{(x-1)}^{2}}{{x}^{2}+2}$ at x=1?

riasc31lj

Beginner2022-04-09Added 8 answers

Explanation:

By definition, the normal line must have a slope that is the opposite reciprocal of the slope of the tangent line at x=1.

Thus, to determine the slope of the normal line, we must first calculate the slope of the tangent line at x=1, which is just the derivative of f at x=1 or f'(1).

Step 1 Calculate$\frac{dy}{dx}$

$\frac{d}{dx}\left[\frac{{(x-1)}^{2}}{{x}^{2}+2}\right]$

$=\frac{2({x}^{2}+2)(x-1)-{(x-1)}^{2}\left(2x\right)}{{({x}^{2}+2)}^{2}}$

Step 2: Find f'(1)

$f}^{\prime}\left(1\right)=\frac{2\left(3\right)\left(0\right)-{\left(0\right)}^{2}\left(2\right)}{{({1}^{2}+2)}^{3}$

f'(1)=0

Step 3: Determine the equation of the normal line

Knowing that f'(1)=0 tells us that there is the graph of f has a horizontal tangent line at x=1. Thus, the normal line must have a slope of$\mathrm{\infty}$ , which is undefined.

Because vertical lines are the only type of line with undefined slopes, f must have a vertical tangent line at x=1.

The equation for the vertical tangent line is x=1.

By definition, the normal line must have a slope that is the opposite reciprocal of the slope of the tangent line at x=1.

Thus, to determine the slope of the normal line, we must first calculate the slope of the tangent line at x=1, which is just the derivative of f at x=1 or f'(1).

Step 1 Calculate

Step 2: Find f'(1)

f'(1)=0

Step 3: Determine the equation of the normal line

Knowing that f'(1)=0 tells us that there is the graph of f has a horizontal tangent line at x=1. Thus, the normal line must have a slope of

Because vertical lines are the only type of line with undefined slopes, f must have a vertical tangent line at x=1.

The equation for the vertical tangent line is x=1.

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