What is the tangent of f(x)= 3/(x^2),\at x=2 ?

The tangent of f(x)= 3/(x^2),\at x=2 is y=-3/4 x+9/4

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tangent of f(x)= 3/(x^2),\at x=2

tangent of

f (x) = \frac{3}{x ^{2}}, at x = 2

y = - \frac{3}{4} x + \frac{9}{4}

Solution steps

Find the tangent point:

(2, \frac{3}{4})

Find the slope of

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

Compute the slope of

f (x) = \frac{3}{x ^{2}}

x = 2 : m = - \frac{3}{4}

Find the line with slope m=

- \frac{3}{4}

and passing through

(2, \frac{3}{4}) : y = - \frac{3}{4} x + \frac{9}{4}

y = - \frac{3}{4} x + \frac{9}{4}

\frac{\partial}{\partial x} (y^{3 x})

f (x) = x + 2, x, x = 1, x = 3

\frac{d}{d x} (\frac{25}{x})

\int \frac{x}{x ^{2} + 5} d x

\frac{d y}{d x} = e^{4 x} + 7 y

What is the tangent of f(x)= 3/(x^2),\at x=2 ?
The tangent of f(x)= 3/(x^2),\at x=2 is y=-3/4 x+9/4