Question

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

Answer 1

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

Explanation:

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.