Absolute Max/Min of f on [0,8]

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Homework Statement

Find the absolute maximum and absolute minimum values of f on the given interval.
f(x) = x/(x^2 + 25)
[0, 8]

Homework Equations

taking derivatives are necessary

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...

 

[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.

 

[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!

 

[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)

 

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

 

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

 

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got it. the answer is correct thanks! :)

 

[jfhorns7853]

My bad, yes f(8) is right.