What is the tangent of f(x)= 8/(x^2),\at x=2 ?

The tangent of f(x)= 8/(x^2),\at x=2 is y=-2x+6

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tangent of f(x)= 8/(x^2),\at x=2

tangent of

f (x) = \frac{8}{x ^{2}}, at x = 2

y = - 2 x + 6

Solution steps

Find the tangent point:

(2, 2)

Find the slope of

f (x) = \frac{8}{x ^{2}} : \frac{df ( x )}{d x} = - \frac{16}{x ^{3}}

Compute the slope of

f (x) = \frac{8}{x ^{2}}

x = 2 : m = - 2

Find the line with slope m=

- 2

and passing through

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

y = - 2 x + 6

x \to 0 lim (\frac{2}{x + 5})

x^{2} + y^{2} = c^{2}

x \to - 2 lim (\frac{x ^{2} + 5 x + 6}{x})

\int \frac{2 x + 18}{x ^{2} + 10 x + 24} d x

y = ln (9 + x^{2})

What is the tangent of f(x)= 8/(x^2),\at x=2 ?
The tangent of f(x)= 8/(x^2),\at x=2 is y=-2x+6