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

The tangent of f(x)=-x^3,\at x=2 is y=-12x+16

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

tangent of

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

y = - 12 x + 16

Solution steps

Find the tangent point:

(2, - 8)

Find the slope of

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

Compute the slope of

f (x) = - x^{3}

x = 2 : m = - 12

Find the line with slope m=

- 12

and passing through

(2, - 8) : y = - 12 x + 16

y = - 12 x + 16

x \to 0 lim (2 x e^{2 x} + k x)

y^{^{'}} = 3 x + y, y (0) = 6

y = ln (e^{- x} + x e^{- x})

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

\int_{1}^{3} \frac{1}{4 + ( x - 1 ) ^{2}} d x

What is the tangent of f(x)=-x^3,\at x=2 ?
The tangent of f(x)=-x^3,\at x=2 is y=-12x+16