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

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

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

tangent of

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

y = 24 x + 32

Solution steps

Find the tangent point:

(- 2, - 16)

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}

x = - 2 : m = 24

Find the line with slope m=

24

and passing through

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

y = 24 x + 32

\int \frac{1}{x 64 x ^{2} - 2} d x

\int \frac{1}{x ^{2} + 16} d x

e^{e^{x}}

\int_{0}^{\frac{π}{4}} sec^{4} (x) d x

ln 3 x^{3} \cdot sec^{3} ({5 x + 4})

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