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{\partial}{\partial x} (\frac{1}{2} sin (2 x - y))

\frac{\partial}{\partial x} (\frac{x ^{3} b ^{4}}{c ^{2} d ^{4}})

y = 4 x, x = 0, y = 4

\frac{d y}{d x} = \frac{( y - x y )}{x ^{2}}

x \to 0 lim (4 - 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