What is the tangent of f(x)=x^3-4x,\at x=-2 ?

The tangent of f(x)=x^3-4x,\at x=-2 is y=8x+16

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tangent of f(x)=x^3-4x,\at x=-2

tangent of

f (x) = x^{3} - 4 x, at x = - 2

y = 8 x + 16

Solution steps

Find the tangent point:

(- 2, 0)

Find the slope of

f (x) = x^{3} - 4 x : \frac{df ( x )}{d x} = 3 x^{2} - 4

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 8

Find the line with slope m=

8

and passing through

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

y = 8 x + 16

\int_{2}^{4} \frac{2 x + 5}{5 - x} d x

y^{^{'}} = x e^{- y}

3 y^{^{''}} + 5 y^{^{'}} - 2 y = 0

x \to 8 - lim (ln (8 - x))

x \to 1 lim (ln (9 x) + e^{x})

What is the tangent of f(x)=x^3-4x,\at x=-2 ?
The tangent of f(x)=x^3-4x,\at x=-2 is y=8x+16