What is the tangent of f(x)=2x^3-6x,\at x=-3 ?

The tangent of f(x)=2x^3-6x,\at x=-3 is y=48x+108

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tangent of f(x)=2x^3-6x,\at x=-3

tangent of

f (x) = 2 x^{3} - 6 x, at x = - 3

y = 48 x + 108

Solution steps

Find the tangent point:

(- 3, - 36)

Find the slope of

f (x) = 2 x^{3} - 6 x : \frac{df ( x )}{d x} = 6 x^{2} - 6

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 = 48

Find the line with slope m=

48

and passing through

(- 3, - 36) : y = 48 x + 108

y = 48 x + 108

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\int \frac{1}{3 - x ^{2} + 4 x} d x

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\frac{d}{d x} (\frac{1}{( x + k ) ^{3}})

\int \frac{x ^{5} + x - 1}{x ^{3} + 1} d x

What is the tangent of f(x)=2x^3-6x,\at x=-3 ?
The tangent of f(x)=2x^3-6x,\at x=-3 is y=48x+108