1. ## min/max

The problem is: Find the absolute maximum and absolute minimum values of f on the given interval.

[-1, 5]

2. Originally Posted by kwivo
The problem is: Find the absolute maximum and absolute minimum values of f on the given interval.

[-1, 5]
first check the endpoints. that is, find f(-1) and f(5).

then search for the critical values. that is, find f'(t) and set it equal to zero and solve for t. then find all f(c) where c represents a critical value, that is, one of the t's you solved for just before. which ever one of these values is the greatest (you include the endpoints as well) is the absolute max, whichever is the lowest is the absolute min

can you continue?

3. I got it. Thanks.

4. Ok actually, I have one more question. I'm not getting the min correct for some reason but I got the max. So, I found the critical point of +or- sqr root of 12.5. So I got the max of 12.5 from f(sqr root of 12.5). Then using the neg of the critical number, I got -12.5, but it's not the right answer...which I don't understand why.

5. Originally Posted by kwivo
Ok actually, I have one more question. I'm not getting the min correct for some reason but I got the max. So, I found the critical point of +or- sqr root of 12.5. So I got the max of 12.5 from f(sqr root of 12.5). Then using the neg of the critical number, I got -12.5, but it's not the right answer...which I don't understand why.
you should find $f(-1),f(5), f \left( \sqrt{\frac {25}2} \right), \mbox{ and }f \left( - \sqrt{\frac {25}2} \right)$, the lowest of these is the absolute min

was that what you did? it seems you plugged -12.5 as opposed to -sqrt(12.5)

try again

6. I did do that. f(-1)=-4.89, f(0)=0

Then I plug in f(-squ root of 12.5), when you do that, t is squared inside the square root. So (-squ root of 12.5)^2 would just give me 12.5. Then 25-12.5=12.5 square root it, times -square root of 12.5 would just get rid of the square root to give me -12.5. That's what I got. What did I do wrong?