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

The tangent of f(x)= 2/(x^2),\at x=1 is y=-4x+6

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

tangent of

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

y = - 4 x + 6

Solution steps

Find the tangent point:

(1, 2)

Find the slope of

f (x) = \frac{2}{x ^{2}} : \frac{df ( x )}{d x} = - \frac{4}{x ^{3}}

Compute the slope of

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

x = 1 : m = - 4

Find the line with slope m=

- 4

and passing through

(1, 2) : y = - 4 x + 6

y = - 4 x + 6

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

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

\frac{d}{d x} ((4 x^{3} - 7 x)^{7})

\int x 2 x - x^{2} d x

\int (\frac{8 + 5 x}{1 + x ^{2}}) d x

What is the tangent of f(x)= 2/(x^2),\at x=1 ?
The tangent of f(x)= 2/(x^2),\at x=1 is y=-4x+6