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

\frac{\partial}{\partial y} (- \frac{5}{1 + x ^{2} - y})

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

\int 21 sec^{4} (x) d x

\frac{d}{d x} (- 3 e^{- x})

x \to 4 lim (3 x^{2} + x + 1)

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