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

The tangent of f(x)=-x^3+3x^2 is y=(-3a_{0}^2+6a_{0})x+2a_{0}^3-3a_{0}^2

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

tangent of

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

y = (- 3 a_{0}^{2} + 6 a_{0}) x + 2 a_{0}^{3} - 3 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} + 3 a_{0}^{2})

Find the slope of

f (x) = - x^{3} + 3 x^{2} : \frac{df ( x )}{d x} = - 3 x^{2} + 6 x

Compute the slope of

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

x = a_{0} : m = - 3 a_{0}^{2} + 6 a_{0}

Find the line with slope m=

- 3 a_{0}^{2} + 6 a_{0}

and passing through

(a_{0}, - a_{0}^{3} + 3 a_{0}^{2}) : y = (- 3 a_{0}^{2} + 6 a_{0}) x + 2 a_{0}^{3} - 3 a_{0}^{2}

y = (- 3 a_{0}^{2} + 6 a_{0}) x + 2 a_{0}^{3} - 3 a_{0}^{2}

f (x) = \frac{4}{( x ^{2} + 1 ) ^{2}}

\frac{d}{d x} (6 x - 9)

y^{^{'''}} - 9 y^{^{''}} + 20 y^{^{'}} = 0

\frac{\partial}{\partial x} (\frac{( x + y )}{x y - 1})

\int_{15}^{30} 9 d x

What is the tangent of f(x)=-x^3+3x^2 ?
The tangent of f(x)=-x^3+3x^2 is y=(-3a_{0}^2+6a_{0})x+2a_{0}^3-3a_{0}^2