What is the tangent of f(x)=-3x^{2/3},\at x=-8 ?

The tangent of f(x)=-3x^{2/3},\at x=-8 is y=-1/((-1)^{1/3)}x-4

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tangent of f(x)=-3x^{2/3},\at x=-8

tangent of

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

y = - \frac{1}{( - 1 ) ^{\frac{1}{3}}} x - 4

Solution steps

Find the tangent point:

(- 8, - 12)

Find the slope of

f (x) = - 3 x^{\frac{2}{3}} : \frac{df ( x )}{d x} = - \frac{2}{x ^{\frac{1}{3}}}

Compute the slope of

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

x = - 8 : m = - \frac{1}{( - 1 ) ^{\frac{1}{3}}}

Find the line with slope m=

- \frac{1}{( - 1 ) ^{\frac{1}{3}}}

and passing through

(- 8, - 12) : y = - \frac{1}{( - 1 ) ^{\frac{1}{3}}} x - 4

y = - \frac{1}{( - 1 ) ^{\frac{1}{3}}} x - 4

x \to + 0 lim (x^{x})

\int \frac{1}{x ( x + 0.5 )} d x

\int y^{- 1} d y

\int_{- 1}^{1} x^{2} - x^{4} d x

\frac{d}{d x} (4 (e^{x})^{9})

What is the tangent of f(x)=-3x^{2/3},\at x=-8 ?
The tangent of f(x)=-3x^{2/3},\at x=-8 is y=-1/((-1)^{1/3)}x-4