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

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

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

tangent of

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

y = (2 - 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}, 2 a_{0} - a_{0}^{2})

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 2 - 2 a_{0}

Find the line with slope m=

2 - 2 a_{0}

and passing through

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

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

\int cos (x) sin^{5} (x) d x

3 - x, - 4 \leq x \leq 2

\frac{d}{d x} ((1 - x^{2})^{\frac{1}{3}})

\frac{2}{x ^{7}}

e^{\frac{- x}{2}}

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