What are the global and local extrema of $f(x)=4x-x^2$ ?

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Answer 1 · 2016-02-12T13:30:14

$x=2$ is both a global and local maximum.

Assume that the domain of $f$ is all real numbers.

Complete the square

$f(x) = -(x-2)^2-4$

As anything squared cannot be less than zero, $f(2)=-4$ is the local maximum as well as the global maximum.

To show using calculus, take the first derivative of $f$ .

$f'(x) = 4-2x$

Solve for $f'(x)=0$ .

$4-2x=0$

$x=2$

Perform the second derivative test.

$f''(x) = -2$

$f''(2) = -2 < 0$

This means that the gradient is decreasing at $x=2$ .

This means that the gradient changes from positive to zero to negative in a small neighborhood around $x=2$ .

This means that $x=2$ is a local maximum.

There are no other local extrenum and $f$ is continuous everywhere.

$x=2$ is also a global maximum.

What are the global and local extrema of f(x)=4x-x^2 f(x)=4x-x^2 ?