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

\frac{\partial}{\partial t} (\frac{( r + s )}{t})

f (x) = 4 x^{3} x^{4}

\frac{d}{d x} (\frac{x sin ( 4 x )}{8})

6 y^{^{''}} + 4 y^{^{'}} + 7 y = 0

f (x) = 2 x^{2} + x - 1

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