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}

\frac{d}{d x} (x^{3} + x + 1)

n = 1 \sum \infty n^{2} \cdot e^{- n}

\int_{1}^{2} 7 4 x^{2} - 3 d x

\int x^{- 10} d x

x \to 2 lim (\frac{- 1}{x - 2})

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