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

The tangent of f(x)=2x^3,\at x=-1 is y=6x+4

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

tangent of

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

y = 6 x + 4

Solution steps

Find the tangent point:

(- 1, - 2)

Find the slope of

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

Compute the slope of

f (x) = 2 x^{3}

x = - 1 : m = 6

Find the line with slope m=

6

and passing through

(- 1, - 2) : y = 6 x + 4

y = 6 x + 4

\int sin (\frac{x}{5}) d x

a x^{3}, at x = - 5

y = 3 arctan (x - 1 + x^{2})

\frac{\partial}{\partial y} (y^{3} cos (x + y))

\int (2 x^{3} + 10 x^{2} - 3 x + 4) d x

What is the tangent of f(x)=2x^3,\at x=-1 ?
The tangent of f(x)=2x^3,\at x=-1 is y=6x+4