# Homework Help: How to find the max and min of a function on a fixed interval.

1. Sep 11, 2008

### wiz0r

Hello, I got the following problem;

Find the min and max of the following function, on the following interval;

f(x) = x^3 - 6x^2 + 9x + 15, [0, 5]

Ok, to my knowledge, what I need to do is

I: Find the first derivative and equal it to zero, so;

f'(x) = 3x^2 - 12x + 9
0 = 3x^2 - 12x + 9

x = {1, 3}

II: Now, I find the second derivative to determine if it's a Minimum or a Maximum, so;

f''(x) = 6x - 12

f''(1) = 6(1) - 12 = -6

Since -6 < 0 it's a minimum

f''(3) = 6(3) - 12 = 6

Since 6 > 0 it's a maximun, right?

Now, what do I do with the interval?? Am I doing it wrong?

Please, help me fast, I got a test in 3 hours, and I need to know this before my test!

Thanks,
~Edwin

2. Sep 11, 2008

### wiz0r

What my friend told me to do is to;

Find the first derivative and it's zeros. Then to evaluate those zeros AND the end points of the interval on the initial equation, and then the biggest value of those will be the maximun and the lowest the minimun (the points), is that right?

3. Sep 11, 2008

### Dick

Yes. Now find f(x) for each of those values (including the endpoints). The smallest of those values is the min, the largest is the max.

4. Sep 11, 2008

### wiz0r

Ok, so;

f(0) = 15
f(1) = 1 - 6 + 9 + 15 = 19
f(3) = 27 - 54 + 27 + 15 = 15
f(5) = 35

Therefore,

Minimum = (0,15) and (3, 15);
Maximum = (5,35)

Am I correct?

5. Sep 11, 2008

### Dick

Looks right to me.

6. Sep 11, 2008

### wiz0r

Woot, thanks a lot!