What is the tangent of f(x)= 1/(x^5),(-3,-1/243) ?

The tangent of f(x)= 1/(x^5),(-3,-1/243) is y=-5/729 x-2/81

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tangent of f(x)= 1/(x^5),(-3,-1/243)

tangent of

f (x) = \frac{1}{x ^{5}}, (- 3, - \frac{1}{243})

y = - \frac{5}{729} x - \frac{2}{81}

Solution steps

Find the slope of

f (x) = \frac{1}{x ^{5}} : \frac{df ( x )}{d x} = - \frac{5}{x ^{6}}

Compute the slope of

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

(- 3, - \frac{1}{243}) : m = - \frac{5}{729}

Find the line with slope m=

- \frac{5}{729}

and passing through

(- 3, - \frac{1}{243}) : y = - \frac{5}{729} x - \frac{2}{81}

y = - \frac{5}{729} x - \frac{2}{81}

\int_{1}^{3} x ln (x) d x

\int \frac{4 - 32 x}{25 - 16 x ^{2}} d x

\frac{d}{d x} (8 + 4 x)

\int x x x d x

f (x) = 8 (x - \frac{1}{x})^{3}, at x = 2

What is the tangent of f(x)= 1/(x^5),(-3,-1/243) ?
The tangent of f(x)= 1/(x^5),(-3,-1/243) is y=-5/729 x-2/81