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

The tangent of f(x)= 1/(x^2) is y=-2/(a_{0)^3}x+3/(a_{0)^2}

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tangent of f(x)= 1/(x^2)

tangent of

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

y = - \frac{2}{a _{0}^{3}} x + \frac{3}{a _{0}^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{1}{a _{0}^{2}})

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 = a_{0} : m = - \frac{2}{a _{0}^{3}}

Find the line with slope m=

- \frac{2}{a _{0}^{3}}

and passing through

(a_{0}, \frac{1}{a _{0}^{2}}) : y = - \frac{2}{a _{0}^{3}} x + \frac{3}{a _{0}^{2}}

y = - \frac{2}{a _{0}^{3}} x + \frac{3}{a _{0}^{2}}

f (x) = x^{4}

x^{11} arccot (x)

(2, - 14), (- 3, 0)

y = 3 x + 5

(2, 4), (2, - 7)

What is the tangent of f(x)= 1/(x^2) ?
The tangent of f(x)= 1/(x^2) is y=-2/(a_{0)^3}x+3/(a_{0)^2}