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

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

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

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

\frac{d}{d x} (\frac{0.6 e ^{x}}{cos ( x )})

x + 3

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