# Absolute max and min

1. Nov 6, 2009

### hover

1. The problem statement, all variables and given/known data
Find the absolute maximum and absolute minimum values of f on the given interval.
f(x) = x/(x^2 + 25)
[0, 8]

2. Relevant equations
taking derivatives are necessary

3. The attempt at a solution

$$f(x)= \frac{x}{x^2+25}$$
$$f'(x)= \frac{(x^2+25)-x(2x)}{(x^2+25)^2}$$
$$f'(x)= \frac{x^2+25-2x^2}{(x^2+25)^2}$$
$$f'(x)= \frac{-x^2+25}{(x^2+25)^2}$$

$$f'(x)= -x^2+25=0$$
$$25=x^2$$
$$x=-5,5$$

$$f(0)=0$$ min
$$f(5)=1/2$$ max
$$f(8)=8/89$$

Those are my answers but i think i am wrong...

2. Nov 6, 2009

### jfhorns7853

You can't cancel
$${-x^2+25}$$
and $${(x^2+25)^2}$$
b/c of the negative.
If you factor out the negative:
$$\frac{-(x^2-25)}{(x^2+25)^2}$$

Which gives you a numerator of
$${-(x+5)(x-5)}$$
which are your zeros of f'(x) and you just plug those x-values of x = +/- 5 into f(x) and find your max and min.

3. Nov 6, 2009

### jfhorns7853

btw, you're x-values are right, but your process is flawed.
Also, b/c of the interval, only f(5) applies from the derivative.

You will also want to graph the equation to make sure.
Graph all of f(x) and then eyeball where 0 and 8 are and just mark through them to show exactly the part of the curve with which you are dealing.
You should do this with all functions, especially rationals. A simple sketch should do the trick most of the time with a few points written in. Don't forget your asymptotes and intercepts too!

Last edited: Nov 6, 2009
4. Nov 6, 2009

### jfhorns7853

Lastly, your f(5) and f(8) values ARE WRONG. Make sure that you plugged 5 and 8 into f(x) and nothing else. Also, upon graphing, you will note that f(8) is not an absolute max or min. Compare f(0), f(5), and f(8)

5. Nov 6, 2009

### hover

f(5)=1/10
but i think f(8) is right

6. Nov 6, 2009

### hover

shouldn't that mean that f(0) is a min and f(5) is a max?

7. Nov 6, 2009

### hover

got it. the answer is correct thanks!! :)

8. Nov 6, 2009

### jfhorns7853

My bad, yes f(8) is right.