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

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

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

tangent of

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

y = 6 x - 6

Solution steps

Find the slope of

y = 3 x^{3} - 3 x : \frac{d y}{d x} = 9 x^{2} - 3

Compute the slope of

y = 3 x^{3} - 3 x

(1, 0) : m = 6

Find the line with slope m=

6

and passing through

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

y = 6 x - 6

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

\frac{\partial}{\partial x} (x 1 - x^{2})

y = 8 x, y = 0, y = x^{2} - 9

\int 4 θ cos (θ) d θ

\int \frac{( 2 + ln ( x ) )}{x} d x

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