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

Nov 17, 2016

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

#### 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 $0$.

$f \left(x\right) = 2 + \left({x}^{2} + 2 x + 1\right) = {x}^{2} + 2 x + 3$

Through the power rule,

$f ' \left(x\right) = 2 x + 2$

Setting equal to $0$,

$2 x + 2 = 0 \text{ "=>" } x = - 1$

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:

$f \left(- 2\right) = 2 + {\left(- 2 + 1\right)}^{2} = 3$

$f \left(- 1\right) = 2 + {\left(- 1 + 1\right)}^{2} = 2$

$f \left(4\right) = 2 + {\left(4 + 1\right)}^{2} = 27$

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