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

The tangent of f(x)=x^3-4x,\at x=0 is y=-4x

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

tangent of

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

y = - 4 x

Solution steps

Find the tangent point:

(0, 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 = 0 : m = - 4

Find the line with slope m=

- 4

and passing through

(0, 0) : y = - 4 x

y = - 4 x

\frac{2}{( s + 2 )}

ln (\frac{( 8 x - 9 ) ^{2}}{- x - 1})

x^{\frac{11}{2}}

x^{^{'}} = - x^{2}

f (2) = 3 x^{2} - 2 x + 5

What is the tangent of f(x)=x^3-4x,\at x=0 ?
The tangent of f(x)=x^3-4x,\at x=0 is y=-4x