What is the tangent of f(x)=2x-x^2,\at x=1 ?

The tangent of f(x)=2x-x^2,\at x=1 is y=1

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tangent of f(x)=2x-x^2,\at x=1

tangent of

f (x) = 2 x - x^{2}, at x = 1

y = 1

Solution steps

Find the tangent point:

(1, 1)

Find the slope of

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

Compute the slope of

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

x = 1 : m = 0

Find the line with slope m=

0

and passing through

(1, 1) : y = 1

y = 1

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

\int cos (x) sin (x) 1 + cos^{2} (x) d x

4 y^{^{''}} - 16 y = 0

\int 4 tan (x) d x

f (x) = 2 e^{x}

What is the tangent of f(x)=2x-x^2,\at x=1 ?
The tangent of f(x)=2x-x^2,\at x=1 is y=1