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

1 Answer
mason m
Dec 18, 2015

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]}

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