zdebe5l8

2022-04-08

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

riasc31lj

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 $\frac{dy}{dx}$
$\frac{d}{dx}\left[\frac{{\left(x-1\right)}^{2}}{{x}^{2}+2}\right]$
$=\frac{2\left({x}^{2}+2\right)\left(x-1\right)-{\left(x-1\right)}^{2}\left(2x\right)}{{\left({x}^{2}+2\right)}^{2}}$
Step 2: Find f'(1)
${f}^{\prime }\left(1\right)=\frac{2\left(3\right)\left(0\right)-{\left(0\right)}^{2}\left(2\right)}{{\left({1}^{2}+2\right)}^{3}}$
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.

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