What is the tangent of f(x)=x^3-4x,(2,0) ?

The tangent of f(x)=x^3-4x,(2,0) is y=8x-16

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tangent of f(x)=x^3-4x,(2,0)

tangent of

f (x) = x^{3} - 4 x, (2, 0)

y = 8 x - 16

Solution steps

Find the slope of

f (x) = x^{3} - 4 x : \frac{df ( x )}{d x} = 3 x^{2} - 4

Compute the slope of

f (x) = x^{3} - 4 x

(2, 0) : m = 8

Find the line with slope m=

8

and passing through

(2, 0) : y = 8 x - 16

y = 8 x - 16

cos (x) - \frac{2}{2} x

x \to 1^{- 1} lim (- 1.6 x - 0.6)

\frac{d}{d x} (cot (3 x^{2} + 5))

π (2 t - 3)

ln (x - 2)

What is the tangent of f(x)=x^3-4x,(2,0) ?
The tangent of f(x)=x^3-4x,(2,0) is y=8x-16