What is the tangent of f(x)= 1/(x^2),(-3, 1/9) ?

The tangent of f(x)= 1/(x^2),(-3, 1/9) is y= 2/27 x+1/3

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tangent of f(x)= 1/(x^2),(-3, 1/9)

tangent of

f (x) = \frac{1}{x ^{2}}, (- 3, \frac{1}{9})

y = \frac{2}{27} x + \frac{1}{3}

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}}

(- 3, \frac{1}{9}) : m = \frac{2}{27}

Find the line with slope m=

\frac{2}{27}

and passing through

(- 3, \frac{1}{9}) : y = \frac{2}{27} x + \frac{1}{3}

y = \frac{2}{27} x + \frac{1}{3}

\int \frac{1}{cos ^{2} ( 3 x )} d x

\int_{- 2}^{2} (x + 3) 4 - x^{2} d x

\int (\frac{x ^{4} + 2 x ^{3} - 3}{x ^{3}}) d x

f (x) = 2 e^{x^{3}} + 1

f (x) = x + 3

What is the tangent of f(x)= 1/(x^2),(-3, 1/9) ?
The tangent of f(x)= 1/(x^2),(-3, 1/9) is y= 2/27 x+1/3