What is the tangent of f(x)=3x-12x^3,\at x=0 ?

The tangent of f(x)=3x-12x^3,\at x=0 is y=3x

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tangent of f(x)=3x-12x^3,\at x=0

tangent of

f (x) = 3 x - 12 x^{3}, at x = 0

y = 3 x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

f (x) = 3 x - 12 x^{3} : \frac{df ( x )}{d x} = 3 - 36 x^{2}

Compute the slope of

f (x) = 3 x - 12 x^{3}

x = 0 : m = 3

Find the line with slope m=

3

and passing through

(0, 0) : y = 3 x

y = 3 x

\int 4 x 1 + 4 x^{2} d x

f (x) = 4 x^{2} e^{2 x^{2} + x - 1}

\int 13 - (\frac{x}{40}) d x

\int x^{2} ln (6 x) d x

What is the tangent of f(x)=3x-12x^3,\at x=0 ?
The tangent of f(x)=3x-12x^3,\at x=0 is y=3x