What is the tangent of f(x)=4x-x^2,\at x=2 ?

The tangent of f(x)=4x-x^2,\at x=2 is y=4

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tangent of f(x)=4x-x^2,\at x=2

tangent of

f (x) = 4 x - x^{2}, at x = 2

y = 4

Solution steps

Find the tangent point:

(2, 4)

Find the slope of

f (x) = 4 x - x^{2} : \frac{df ( x )}{d x} = 4 - 2 x

Compute the slope of

f (x) = 4 x - x^{2}

x = 2 : m = 0

Find the line with slope m=

0

and passing through

(2, 4) : y = 4

y = 4

s = t^{3} - t^{2}

\frac{d}{d x} (16 x - x^{2})

t^{2} y^{^{''}} + 7 t y^{^{'}} + 9 y = 0

\int \frac{x ^{5}}{( x ^{3} + 1 ) ^{2}} d x

x \to 1 lim (cos (\frac{π}{2} x))

What is the tangent of f(x)=4x-x^2,\at x=2 ?
The tangent of f(x)=4x-x^2,\at x=2 is y=4