What is the tangent of f(x)= 1/(x^3),\at x=5 ?

The tangent of f(x)= 1/(x^3),\at x=5 is y=-3/625 x+4/125

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tangent of f(x)= 1/(x^3),\at x=5

tangent of

f (x) = \frac{1}{x ^{3}}, at x = 5

y = - \frac{3}{625} x + \frac{4}{125}

Solution steps

Find the tangent point:

(5, \frac{1}{125})

Find the slope of

f (x) = \frac{1}{x ^{3}} : \frac{df ( x )}{d x} = - \frac{3}{x ^{4}}

Compute the slope of

f (x) = \frac{1}{x ^{3}}

x = 5 : m = - \frac{3}{625}

Find the line with slope m=

- \frac{3}{625}

and passing through

(5, \frac{1}{125}) : y = - \frac{3}{625} x + \frac{4}{125}

y = - \frac{3}{625} x + \frac{4}{125}

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What is the tangent of f(x)= 1/(x^3),\at x=5 ?
The tangent of f(x)= 1/(x^3),\at x=5 is y=-3/625 x+4/125