What is the tangent of f(x)=x^3-5x^2,\at x=0 ?

The tangent of f(x)=x^3-5x^2,\at x=0 is y=0

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tangent of f(x)=x^3-5x^2,\at x=0

tangent of

f (x) = x^{3} - 5 x^{2}, at x = 0

y = 0

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

f (x) = x^{3} - 5 x^{2} : \frac{df ( x )}{d x} = 3 x^{2} - 10 x

Compute the slope of

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

x = 0 : m = 0

Find the line with slope m=

0

and passing through

(0, 0) : y = 0

y = 0

x \to 5 lim (3 x^{2} + 3 x + 5)

\int \frac{1}{n} d n

n = 1 \sum \infty \frac{n}{e ^{n}}

\int arctan (2 x + 1) d x

\int 3 e^{4 x} d x

What is the tangent of f(x)=x^3-5x^2,\at x=0 ?
The tangent of f(x)=x^3-5x^2,\at x=0 is y=0