Help with finding absolute max/min values for a function

[denbal87]

Hey guys,

Here's the problem:

Find the absolute maximum and absolute minimum values of f on the given interval. x/(x^2 - x + 16), [0, 12]

Here's what I've done so far:

f'(x) = x^2 - x + 16 - x(2x -1)/(x^2 - x + 16)^2 = (x^2 -2x^2 + 16)/(x^2 - x + 16)^2 =
= (16 - x^2)/(x^2 - x + 16)^2
critical numbers: x = 4/-4

f(4) = 4/ 16 - 4 + 16 = 4/28 = 1/6

end points:
f(0) = 0
f(12) = 12/(144 - 12 +16) = 6/74 = 3/37

None of these values seem to be right. Any ideas are appreciated. Thanks!

 

[Ishuda]

Hey guys,

Here's the problem:

Find the absolute maximum and absolute minimum values of f on the given interval. x/(x^2 - x + 16), [0, 12]

Here's what I've done so far:

f'(x) = x^2 - x + 16 - x(2x -1)/(x^2 - x + 16)^2 = (x^2 -2x^2 + 16)/(x^2 - x + 16)^2 =
= (16 - x^2)/(x^2 - x + 16)^2
critical numbers: x = 4/-4

f(4) = 4/ 16 - 4 + 16 = 4/28 = 1/7

end points:
f(0) = 0
f(12) = 12/(144 - 12 +16) = 6/74 = 3/37

None of these values seem to be right. Any ideas are appreciated. Thanks!

Just because I got the same answers (almost, see red correction) doesn't mean the answers are right but I'm pretty sure they are.

 

[denbal87]

Just because I got the same answers (almost, see red correction) doesn't mean the answers are right but I'm pretty sure they are.

Thanks! 1/7 is the max. But why is f(0) the min and not f(-4)?

 

[Ishuda]

Thanks! 1/7 is the max. But why is f(0) the min and not f(-4)?

Because the problem restricts x to the closed interval [0, 12]. Had the interval been, for example, [-12, 12] then f(-4) would have been the correct answer for the minimum.

 

[denbal87]

Because the problem restricts x to the closed interval [0, 12]. Had the interval been, for example, [-12, 12] then f(-4) would have been the correct answer for the minimum.

Ah, of course! Thanks, Ishuda!