What is the tangent of f(x)=x^3-4x^2 ?

The tangent of f(x)=x^3-4x^2 is y=(3a_{0}^2-8a_{0})x-2a_{0}^3+4a_{0}^2

Popular >

tangent of f(x)=x^3-4x^2

tangent of

f (x) = x^{3} - 4 x^{2}

y = (3 a_{0}^{2} - 8 a_{0}) x - 2 a_{0}^{3} + 4 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} - 4 a_{0}^{2})

Find the slope of

f (x) = x^{3} - 4 x^{2} : \frac{df ( x )}{d x} = 3 x^{2} - 8 x

Compute the slope of

f (x) = x^{3} - 4 x^{2}

x = a_{0} : m = 3 a_{0}^{2} - 8 a_{0}

Find the line with slope m=

3 a_{0}^{2} - 8 a_{0}

and passing through

(a_{0}, a_{0}^{3} - 4 a_{0}^{2}) : y = (3 a_{0}^{2} - 8 a_{0}) x - 2 a_{0}^{3} + 4 a_{0}^{2}

y = (3 a_{0}^{2} - 8 a_{0}) x - 2 a_{0}^{3} + 4 a_{0}^{2}

\int_{0}^{\frac{π}{4}} sin (x) d x

\frac{48}{( 1 - x ) ^{4}}

\frac{d ^{2}}{d x ^{2}} ((2 x^{2} + 7)^{5})

f (x) = cos^{2} (x) + sin^{2} (x)

y^{^{''}} - y^{^{'}} - 2 y = - 8 t + 8 t^{2}

What is the tangent of f(x)=x^3-4x^2 ?
The tangent of f(x)=x^3-4x^2 is y=(3a_{0}^2-8a_{0})x-2a_{0}^3+4a_{0}^2