What is the tangent of f(x)=2x^3,(-2,-16) ?

The tangent of f(x)=2x^3,(-2,-16) is y=24x+32

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tangent of f(x)=2x^3,(-2,-16)

tangent of

f (x) = 2 x^{3}, (- 2, - 16)

y = 24 x + 32

Solution steps

Find the slope of

f (x) = 2 x^{3} : \frac{df ( x )}{d x} = 6 x^{2}

Compute the slope of

f (x) = 2 x^{3}

(- 2, - 16) : m = 24

Find the line with slope m=

24

and passing through

(- 2, - 16) : y = 24 x + 32

y = 24 x + 32

\int cot (3 x) d x

ln (x x^{2} - 5)

\frac{7}{1 - x}

\frac{d}{d x} (2 e^{- x} + x e^{- x})

x^{2} + x + 8

What is the tangent of f(x)=2x^3,(-2,-16) ?
The tangent of f(x)=2x^3,(-2,-16) is y=24x+32