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

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

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

tangent of

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

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

Find the slope of

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

Compute the slope of

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

x = a_{0} : m = 3 a_{0}^{2} - 4 a_{0}

Find the line with slope m=

3 a_{0}^{2} - 4 a_{0}

and passing through

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

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

x \to \infty lim (- 5 x)

x^{4} + y^{3} x^{5} + 4 y^{^{'}} = 0, y (0) = 0

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

7 (e^{x} cos (x) - e^{x} sin (x))

\frac{d}{d x} (x - 1 + \frac{16}{x})

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