What is the tangent of 4x-x^2 ?

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

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tangent of 4x-x^2

tangent of

4 x - x^{2}

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 a_{0} - a_{0}^{2})

Find the slope of

y = 4 x - x^{2} : \frac{d y}{d x} = 4 - 2 x

Compute the slope of

y = 4 x - x^{2}

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

Find the line with slope m=

4 - 2 a_{0}

and passing through

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

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

ln (x + y)

\frac{\partial}{\partial z} (cos (z w (z)))

x \to 0 lim (\frac{4 ^{x} - 7 ^{x}}{x})

\frac{\partial}{\partial x} (e^{- 6 x} cos (2 π t))

\int (3 x - 2 x^{- 3}) d x