What is the tangent of f(x)=3x^3-3x,(-1,0) ?

The tangent of f(x)=3x^3-3x,(-1,0) is y=6x+6

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tangent of f(x)=3x^3-3x,(-1,0)

tangent of

f (x) = 3 x^{3} - 3 x, (- 1, 0)

y = 6 x + 6

Solution steps

Find the slope of

f (x) = 3 x^{3} - 3 x : \frac{df ( x )}{d x} = 9 x^{2} - 3

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 = 6

Find the line with slope m=

6

and passing through

(- 1, 0) : y = 6 x + 6

y = 6 x + 6

\frac{\partial}{\partial u} (4 - u^{2} - v^{2})

\int - 2 x - 1 d x

x \to \infty lim (0 \cdot x)

\int tan (x) sec^{\frac{3}{2}} (x) d x

\frac{d}{d θ} (cos (4 θ))

What is the tangent of f(x)=3x^3-3x,(-1,0) ?
The tangent of f(x)=3x^3-3x,(-1,0) is y=6x+6