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}}

Compute the slope of

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

(- 1, 1) : 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}

y^{^{'}} = 0.3 y (1 - \frac{y}{2500}) - 120

\frac{6 s + 3}{s ^{4} + 5 s ^{2} + 4}

x y^{^{'}} + 4 y = 8 x^{4}

\int e^{- t} t d t

\int 27 x^{7} + 3 x^{5} - 45 x^{3} + 2 x d x

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