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

\frac{d}{d z} (\frac{z ^{2}}{( z - 2 i ) ^{2}})

\frac{d}{d x} ((1 + 2 sin (x)) sin (x))

\int \frac{x arctan ( x )}{( 1 + x ^{2} ) ^{2}} d x

\int_{0}^{1} 2 π (x + 8) x^{2} d x

x \to - 2 lim (- 3 x^{2} - 6 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