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

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

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

tangent of

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

y = 8 x - 16

Solution steps

Find the tangent point:

(2, 0)

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

x = 2 : m = 8

Find the line with slope m=

8

and passing through

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

y = 8 x - 16

2 x^{2} - 1, 0, 1

1 - 4 x^{2}

\frac{x + 5}{x}

f (x) = x^{2} - 9 x, at x = 3

\frac{\partial}{\partial y} (x^{3} y)

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