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

The tangent of f(x)=x^3-5x^2 is y=(3a_{0}^2-10a_{0})x-2a_{0}^3+5a_{0}^2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=x^3-5x^2

tangent of

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

y = (3 a_{0}^{2} - 10 a_{0}) x - 2 a_{0}^{3} + 5 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3} - 5 a_{0}^{2})

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 = a_{0} : m = 3 a_{0}^{2} - 10 a_{0}

Find the line with slope m=

3 a_{0}^{2} - 10 a_{0}

and passing through

(a_{0}, a_{0}^{3} - 5 a_{0}^{2}) : y = (3 a_{0}^{2} - 10 a_{0}) x - 2 a_{0}^{3} + 5 a_{0}^{2}

y = (3 a_{0}^{2} - 10 a_{0}) x - 2 a_{0}^{3} + 5 a_{0}^{2}

\int_{- 2}^{0} x^{2} e^{x^{3} + 8} d x

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

\int (x - 1) (x + 1) d x

\int x^{3} (2 x^{4} + 4)^{2} d x

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

What is the tangent of f(x)=x^3-5x^2 ?
The tangent of f(x)=x^3-5x^2 is y=(3a_{0}^2-10a_{0})x-2a_{0}^3+5a_{0}^2