What is the tangent of f(x)=2x^3-8x,(-2,0) ?

The tangent of f(x)=2x^3-8x,(-2,0) is y=16x+32

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tangent of f(x)=2x^3-8x,(-2,0)

tangent of

f (x) = 2 x^{3} - 8 x, (- 2, 0)

y = 16 x + 32

Solution steps

Find the slope of

f (x) = 2 x^{3} - 8 x : \frac{df ( x )}{d x} = 6 x^{2} - 8

Compute the slope of

f (x) = 2 x^{3} - 8 x

(- 2, 0) : m = 16

Find the line with slope m=

16

and passing through

(- 2, 0) : y = 16 x + 32

y = 16 x + 32

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

f (x) = x \cdot sin (x)

\frac{d}{d x} (\frac{8 x ^{2} + 2 x + 8}{x})

x \to + 0 lim (x sin (\frac{3}{x}))

\frac{d}{d x} (\frac{3 - \frac{1}{x}}{x + 5})

What is the tangent of f(x)=2x^3-8x,(-2,0) ?
The tangent of f(x)=2x^3-8x,(-2,0) is y=16x+32