Question

What are the extrema of `f(x) = 2 + (x + 1)^2` on #[-2,4]?

Answer 1

There is a global minimum of `2` at `x=-1` and a global maximum of `27` at `x=4` on the interval `[-2,4]`.

Explanation:

Global extrema could occur on an interval at one of two places: at an endpoint or at a critical point within the interval. The endpoints, which we will have to test, are `x=-2` and `x=4`.

To find any critical points, find the derivative and set it equal to `0`.

`f(x)=2+(x^2+2x+1)=x^2+2x+3`

Through the power rule,

`f'(x)=2x+2`

Setting equal to `0`,

`2x+2=0" "=>" "x=-1`

There is a critical point at `x=-1`, which means it could also be a global extremum.

Test the three points we've found to find the maximum and minimum for the interval:

`f(-2)=2+(-2+1)^2=3`

`f(-1)=2+(-1+1)^2=2`

`f(4)=2+(4+1)^2=27`

Thus there is a global minimum of `2` at `x=-1` and a global maximum of `27` at `x=4` on the interval `[-2,4]`.