1. absolute maximum and mins

Find the absolute maximum and absolute minimum values of f on the given interval.

f(x)= x/((x^2) +4)
[0, 3]

I get my derivative, set it to zero and solve but im getting the wrong answer. any ideas?

thanks,
Jay

2. Originally Posted by jfisk87
Find the absolute maximum and absolute minimum values of f on the given interval.

f(x)= x/((x^2) +4)
[0, 3]

I get my derivative, set it to zero and solve but im getting the wrong answer. any ideas?

thanks,
Jay
find the second derivative and solve.

3. You are probably taking the first derivative incorrectly.

4. Originally Posted by jfisk87
Find the absolute maximum and absolute minimum values of f on the given interval.

f(x)= x/((x^2) +4)
[0, 3]

I get my derivative, set it to zero and solve but im getting the wrong answer. any ideas?

thanks,
Jay
Show us what derivative you got, and what answer you got and we will be able to help you better. Note that while max and min may be at a point where the derivative is 0, absolute max and min may also be at the endpoints of the interval. Did you check the value at the endpoints?

(I can see no reason at all for taking the second derivative.)

5. f(x) = x/(x^2+4)

Quotient Rule:
(gf'-fg')/g^2
f=x
g=x^2+4
f'=1
g'=2x

[(x^2+4)1-x(2x)]/(x^2+4)^2
(-x^2+4)/(x^2+4)^2=-x^2+4

0 = plus or minus 2
2 is in the period.
Therefore, you test on either side of 2.
I chose 1 and 3 for test values.
At 1, the slope of the tangent line to the function is positive, meaning that the function is increasing there. (You can draw this out and see it. You might already know that. :P)
At 3, f'(x) (the slope of the tangent line to the function) is negative, meaning that the function is decreasing there.

Therefore, there is a max at 2. There is no min.
If you plug 2 back into the original function...
y = x/(x^2+4)
y = x/(x^2+4) = 2/8 = .25

Therefore, there is a min in the period at (2, .25).

6. Originally Posted by HallsofIvy
Show us what derivative you got, and what answer you got and we will be able to help you better. Note that while max and min may be at a point where the derivative is 0, absolute max and min may also be at the endpoints of the interval. Did you check the value at the endpoints?

(I can see no reason at all for taking the second derivative.)
hi "HallsofIvy"
there's a global maximum at x=2.
if u solve the first derivative you'll find that x=2 is a solution.
sorry,a typo