How do you find the exact relative maximum and minimum of the polynomial function of $f(x)=4x-x^2$ ?

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Answer 1 · 2016-08-18T05:41:16

At $x=2$

$dy/dx=0$ and $(d^2y)/(dx^2)<0$

Hence the function has a maximum at $x=2$

$y=4x-x^2$

$dy/dx=4-2x$

$dy/dx=>4-2x=0$

$x=(-4)/(-2)=2$

$(d^2y)/(dx^2)=-2$

At $x=2$

$dy/dx=0$ and $(d^2y)/(dx^2)<0$

Hence the function has a maximum at $x=2$

graph{4x-x^2 [-10, 10, -5, 5]}

How do you find the exact relative maximum and minimum of the polynomial function of f(x)=4x-x^2f(x)=4x-x^2?