What is the tangent of f(x)=2x^3-2x,(1,0) ?

The tangent of f(x)=2x^3-2x,(1,0) is y=4x-4

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tangent of f(x)=2x^3-2x,(1,0)

tangent of

f (x) = 2 x^{3} - 2 x, (1, 0)

y = 4 x - 4

Solution steps

Find the slope of

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

Compute the slope of

f (x) = 2 x^{3} - 2 x

(1, 0) : m = 4

Find the line with slope m=

4

and passing through

(1, 0) : y = 4 x - 4

y = 4 x - 4

\frac{\partial}{\partial x} (x e^{x + y})

\int 2 arctan (3 y) d y

\frac{\partial}{\partial x} (9 + x ln (x y - 5))

\int_{0}^{\infty} sin (9 x) d x

\frac{d}{d x} (sin (x) - x cos (x))

What is the tangent of f(x)=2x^3-2x,(1,0) ?
The tangent of f(x)=2x^3-2x,(1,0) is y=4x-4