What is the equation of the line normal to f(x)=x−1x2−2 at x=1?

Explanation:At x=1, f(x)=x−1x2−2=1−112−2=0So normal or tangent will be at x=1 will be at (1,0) on the curve.Slope of tangent is given by f'(x) which is given byf′(x)=(x2−2)ddx(x−1)−(x−1)ddx(x2−2)(x2−2)2 or=(x2−2)×1−(x−1)×2x(x2−2)2=x2−2−2x2+2x(x2−2)2 or=−x2−2x+2(x2−2)2Hence, at x=1,f′(X)=−12−2⋅1+2(1−2)2=−1As slope of tangent is -1, slope of normal would be −1−1=1Hence using point slope form equation of normal is(y-0)=1(x-1) or x-y-1=0

Solved: "What is the equation of the line normal..."

Vanessa Mccarty

Answered question

2022-04-09

What is the equation of the line normal to

f (x) = \frac{x - 1}{x^{2} - 2}

at x=1?

Answer & Explanation

Vegljamzt6

Beginner2022-04-10Added 16 answers

Explanation:
At x=1,

f (x) = \frac{x - 1}{x^{2} - 2} = \frac{1 - 1}{1^{2} - 2} = 0

So normal or tangent will be at x=1 will be at (1,0) on the curve.
Slope of tangent is given by f'(x) which is given by

f^{'} (x) = \frac{(x^{2} - 2) \frac{d}{d x} (x - 1) - (x - 1) \frac{d}{d x} (x^{2} - 2)}{{(x^{2} - 2)}^{2}}

= \frac{(x^{2} - 2) \times 1 - (x - 1) \times 2 x}{{(x^{2} - 2)}^{2}} = \frac{x^{2} - 2 - 2 x^{2} + 2 x}{{(x^{2} - 2)}^{2}}

= - \frac{x^{2} - 2 x + 2}{{(x^{2} - 2)}^{2}}

Hence, at

x = 1, f^{'} (X) = - \frac{1^{2} - 2 \cdot 1 + 2}{{(1 - 2)}^{2}} = - 1

As slope of tangent is -1, slope of normal would be

- \frac{1}{- 1} = 1

Hence using point slope form equation of normal is
(y-0)=1(x-1) or x-y-1=0

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