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

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

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tangent of f(x)= 1/(x^2),\at x=2

tangent of

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

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

Solution steps

Find the tangent point:

(2, \frac{1}{4})

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}}

x = 2 : 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}

x y^{^{'}} = y + x^{2} sin (x), y (\frac{π}{3}) = 0

h (31 - \frac{1}{2} h)^{2}

\int_{0}^{1} \frac{sin ( x )}{x} d x

\int 4 ln (4 x^{2}) d x

\frac{d}{d x} (\frac{a ^{2} + x ^{2}}{a ^{2} - x ^{2}})

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