What is the tangent of f(x)= 1/(x^2),(-2, 1/4) ?

The tangent of f(x)= 1/(x^2),(-2, 1/4) is y= 1/4 x+3/4

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tangent of f(x)= 1/(x^2),(-2, 1/4)

tangent of

f (x) = \frac{1}{x ^{2}}, (- 2, \frac{1}{4})

y = \frac{1}{4} x + \frac{3}{4}

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}}

(- 2, \frac{1}{4}) : m = \frac{1}{4}

Find the line with slope m=

\frac{1}{4}

and passing through

(- 2, \frac{1}{4}) : y = \frac{1}{4} x + \frac{3}{4}

y = \frac{1}{4} x + \frac{3}{4}

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\int 5 \frac{x}{16 - x ^{2}} d x

\frac{d}{d x} ((x + 3)^{3})

\int - 3 x cos (4 x) d x

\int (- 5 x^{4} + 8 x^{3}) (- 2 x^{5} + 4 x^{4})^{6} d x

What is the tangent of f(x)= 1/(x^2),(-2, 1/4) ?
The tangent of f(x)= 1/(x^2),(-2, 1/4) is y= 1/4 x+3/4