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

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

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

tangent of

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

y = - 6 x + 4

Solution steps

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}

(1, - 2) : m = - 6

Find the line with slope m=

- 6

and passing through

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

y = - 6 x + 4

\frac{d}{d x} (x (x + 12))

\frac{\partial}{\partial x} (3 z cos (4 x y) + x - 7 y)

f (x) = \frac{1}{6 x ^{3}}

f (x) = \frac{2}{x - 6}, at x = 8

\int_{2}^{5} \frac{x}{x ^{2} + 4 x + 20} d x

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