Optimization

Apr 2010
156
0
When doing an optimization problem, when would you use the closed interval method to get your values? Also if you obtain a critical point where the first derivative is 0, how would you know if this is a max or min?
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
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.
 
Apr 2010
156
0
I know that you had to do that but like would you have to test values with a sign table to the left, between, and right of your critical points with the original function that you differentiated?

Also I know that if f''(x) is + then the graph is concave up and if f''(x) is - then concave down, but I dont understand this in terms of the 2nd derivative test graphically. Must you just memorize this rule and say if critical point + it min, and if - then max?