What are the extrema of # f(x)=x/(x^2+9)# on the interval [0,5]?

Question

Elijah Abram · Accepted Answer

Find the critical values of #f(x)# on the interval #[0,5]#. #f'(x)=((x^2+9)d/dx[x]-xd/dx[x^2+9])/(x^2+9)^2# #f'(x)=(x^2+9-2x^2)/(x^2+9)^2# #f'(x)=-(x^2-9)/(x^2+9)^2# #f'(x)=0# when #x=+-3#.
#f'(x)# is never undefined. To find the extrema, plug in the endpoints of the interval and any critical numbers inside the interval into #f(x)#, which, in this case, is only #3#. #f(0)=0larr"absolute minimum"# #f(3)=1/6larr"absolute maximum"# #f(5)=5/36# Check a graph: graph{x/(x^2+9) [-0.02, 5, -0.02, 0.2]}

Scarlett Allison · Answer

undefined To find the extrema of $ f(x) = \frac{x}{x^2 + 9} $ on the interval $[0, 5]$, we first need to find the critical points by setting the derivative of $ f(x) $ equal to zero and then checking the endpoints of the interval.

1. Find the derivative of $ f(x) $:
$$ f(x) = \frac{x}{x^2 + 9} $$
$$ f'(x) = \frac{(x^2 + 9) - x(2x)}{(x^2 + 9)^2} $$
$$ f'(x) = \frac{9 - x^2}{(x^2 + 9)^2} $$

2. Set $ f'(x) = 0 $ to find critical points:
$$ \frac{9 - x^2}{(x^2 + 9)^2} = 0 $$
$$ 9 - x^2 = 0 $$
$$ x^2 = 9 $$
$$ x = \pm 3 $$

The critical points are $ x = 3 $ and $ x = -3 $. However, $ x = -3 $ is not in the interval $[0, 5]$, so we only consider $ x = 3 $.

3. Evaluate $ f(x) $ at the critical point and endpoints:
- At $ x = 0 $:
$$ f(0) = \frac{0}{0^2 + 9} = 0 $$
- At $ x = 5 $:
$$ f(5) = \frac{5}{5^2 + 9} = \frac{5}{34} $$
- At $ x = 3 $ (critical point):
$$ f(3) = \frac{3}{3^2 + 9} = \frac{3}{18} = \frac{1}{6} $$

The function $ f(x) $ has extrema at $ x = 0 $ and $ x = 3 $ on the interval $[0, 5]$, with $ x = 3 $ being the minimum and $ x = 0 $ being the maximum.