What is the tangent of f(x)= 1/(x^3),(2, 1/8) ?

The tangent of f(x)= 1/(x^3),(2, 1/8) is y=-3/16 x+1/2

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tangent of f(x)= 1/(x^3),(2, 1/8)

tangent of

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

y = - \frac{3}{16} x + \frac{1}{2}

Solution steps

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - \frac{3}{16}

Find the line with slope m=

- \frac{3}{16}

and passing through

(2, \frac{1}{8}) : y = - \frac{3}{16} x + \frac{1}{2}

y = - \frac{3}{16} x + \frac{1}{2}

\frac{d}{d x} (\frac{1}{x ^{2} + y ^{2} + z ^{2}})

\int e^{- x} cos (9 x) d x

\frac{d}{d x} (x (5 x + 9)^{\frac{1}{3}})

\int_{0}^{1} 6 cos (\frac{π t}{2}) d t

\frac{d}{d x} ((4 - x)^{2})

What is the tangent of f(x)= 1/(x^3),(2, 1/8) ?
The tangent of f(x)= 1/(x^3),(2, 1/8) is y=-3/16 x+1/2