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 x sin (3 x - 2) d x

- cos (2 t)

\int 21 x sec^{2} (x) d x

\int \frac{1}{x ln ( x ) ln ( ln ( x ) )} d x

f (x) = 1 1^{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