What is the tangent of f(x)= 6/(x^2),\at x=1 ?

The tangent of f(x)= 6/(x^2),\at x=1 is y=-12x+18

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tangent of f(x)= 6/(x^2),\at x=1

tangent of

f (x) = \frac{6}{x ^{2}}, at x = 1

y = - 12 x + 18

Solution steps

Find the tangent point:

(1, 6)

Find the slope of

f (x) = \frac{6}{x ^{2}} : \frac{df ( x )}{d x} = - \frac{12}{x ^{3}}

Compute the slope of

f (x) = \frac{6}{x ^{2}}

x = 1 : m = - 12

Find the line with slope m=

- 12

and passing through

(1, 6) : y = - 12 x + 18

y = - 12 x + 18

y^{^{''}} - 3 y^{^{'}} - 10 y = 0

\frac{1}{( 1 + x ) ^{2}}, 1

\frac{d}{d x} (5 x)

\int 4 x^{2} + 1 d x

5 x^{2} + 2 x + 3

What is the tangent of f(x)= 6/(x^2),\at x=1 ?
The tangent of f(x)= 6/(x^2),\at x=1 is y=-12x+18