How do you find the extrema for f(x) = x^2 +2x - 4 for [-1,1]?

Aug 9, 2015

Function $f \left(x\right) = {x}^{2} + 2 x - 4$ has in $\left[- 1 , 1\right]$ a minimum $- 5$ for $x = - 1$.

Explanation:

For any function you can use the first derivative - only a point that suffices $f ' \left(x\right) = 0$ can be an extremum. If you're looking for extrema within a given interval you work only with those almost-extrema in the integral.

$f \left(x\right) = {x}^{2} + 2 x - 4$
$f ' \left(x\right) = 2 x + 2$
$2 x + 2 = 0 \implies x = - 1 \in \left[- 1 , 1\right]$

In this case the only point which can be considered in finding the extrema is $x = - 1$ and fortunately it's in the given interval. Now, we evaluate the second derivative and
if $f ' ' \left(x\right) > 0$ then $x$ is a local minimum
if $f ' ' \left(x\right) < 0$ then $x$ is a local maximum.

$f ' ' \left(x\right) = 2$
$f ' ' \left(- 1\right) = 2 > 0$ - $x = - 1$ is a local minimum and f(-1)=-5#.