What is the tangent of 4x-x^2,(1.3) ?

The tangent of 4x-x^2,(1.3) is y=(4-2a_{0})x+a_{0}^2

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tangent of 4x-x^2,(1.3)

tangent of

4 x - x^{2}, (1.3)

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}

f (x) = 5 x^{4} - 6 x^{2} + 1

\frac{\partial}{\partial x} (2 x y - 3)

\frac{\partial}{\partial x} (\frac{4}{( x + 2 ) ^{2}})

\frac{4}{x}, 3

t \to 1 lim (\frac{t ^{2} - 1}{t - 1})

What is the tangent of 4x-x^2,(1.3) ?
The tangent of 4x-x^2,(1.3) is y=(4-2a_{0})x+a_{0}^2