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

Compute the slope of

f (x) = 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

\int sin (2) (π x) cos (5) (π x) d x

y^{^{''}} - 2 y^{^{'}} + 1 y = 0

\int \frac{sin ^{3} ( θ )}{cos ^{6} ( θ )} d θ

\int \frac{sin ( 2 x )}{2} d x

\frac{\partial}{\partial x} (x^{z} y^{1 - z})

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