What is the normal of f(x)= 1/(x^2),(-1,1) ?

The normal of f(x)= 1/(x^2),(-1,1) is f(x)=-1/2 x+1/2

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normal of f(x)= 1/(x^2),(-1,1)

normal of

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

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

Solution steps

Find the slope of

f (x) = \frac{1}{x ^{2}} : \frac{df ( x )}{d x} = - \frac{2}{x ^{3}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 2

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{2}

Find the line with slope m=

- \frac{1}{2}

and passing through

(- 1, 1) : f (x) = - \frac{1}{2} x + \frac{1}{2}

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

What is the normal of f(x)= 1/(x^2),(-1,1) ?
The normal of f(x)= 1/(x^2),(-1,1) is f(x)=-1/2 x+1/2