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

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

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

tangent of

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

y = 0

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

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

Compute the slope of

f (x) = x^{3}

x = 0 : m = 0

Find the line with slope m=

0

and passing through

(0, 0) : y = 0

y = 0

f (x) = \frac{3 x ^{2} - 18 x + 15}{5}

\frac{d}{d x} (\frac{3 sin ( x )}{3 + cos ( x )})

\frac{\partial}{\partial y} (2 x^{3} y + 4)

\int (x^{\frac{3}{2}} + 1)^{2} d x

f (x) = sin (sin (sin (x)))

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