When doing an optimization problem, when would you use the closed interval method to get your values?

Only if the problem were

**given** on a closed interval! Of course, if the interval is not closed and bounded, there might not be a "max" or "min".

Also if you obtain a critical point where the first derivative is 0, how would you know if this is a max or min?

You read in the next couple of pages of your text book about the first and second derivative tests!

**first derivative test** If \(\displaystyle x_0\) is a critical point and f'(x)< 0 for \(\displaystyle x<x_0\) and f'(x)> 0 for \(\displaystyle x> x_0\), \(\displaystyle f(x_0)\) is a

**minimum**.

(Think about what f'> 0 and f'< 0 mean.)

For example, \(\displaystyle f(x)= x^2\) has \(\displaystyle f'(x)= 2x\) and so has x= 0 as a critical point. For x< 0, f'(x)= 2x< 0 while for x> 0, f'(x)= 2x> 0 so x= 0 gives f(0)= 0, a minimum.

If \(\displaystyle x_0\) is a critical point and f'(x)> 0 for \(\displaystyle x<x_0\) and f'(x)> 0 for \(\displaystyle x< x_0\), \(\displaystyle f(x_0)\) is a

**maximum**.

For example, \(\displaystyle f(x)= -x^2\) has \(\displaystyle f'(x)= -2x\) and so has x= 0 as a critical point. For x< 0, f'(x)= -2x> 0 while for x> 0, f'(x)= -2x0 0 so x= 0 gives f(0)= 0, a maximum.

**second derivative test** If \(\displaystyle x_0\) is a critical point and \(\displaystyle f''(x_0)> 0\) then \(\displaystyle f(x_0)\) is a minimum.

For example, for \(\displaystyle f(x)= x^2\), f''(x)= 2> 0 so the critical point x= 0 gives a minimum value.

If \(\displaystyle x_0\) is a critical point and f''< 0[/tex] then \(\displaystyle f(x_0)\) is a maximum.

for example, for \(\displaystyle f(x)= -2x^2\), f''(x)= -2< 0 so the crtical point x= 0 gives a maximum value.

Of course, it might happen that none of those conditions hold- that is that f' does not change sign or that f''= 0. In that case, it might be that the critical point is

**neither** a maximum nor a minimum or it might be that you need some other test.