Asked by Chloe

Find the absolute maximum value and the absolute minimum value, if any, of the function. (If an answer does not exist, enter DNE.)
g(x) = x sqrt(9 − x2) on [0, 3]
maximum:
minimum:
Answered by oobleck
g(x) = √(9-x^2)
g'(x) = -x/√(9-x^2)
You know that on [0,3] this is just a 1/4 circle of radius 3, so there is no local max or min.
abs max = 3
abs min = 0

Don't get caught up in the calculus, and forget your Algebra I.
Answered by Chloe
max as 3 is not correct
Answered by oobleck
rubbish
√(9-0^2) = 3
Is the interval closed? If it supposed to be (0,3] then DNE would be correct.
The limit of g(x) is 3, but there is no "largest" number less than 3.
Answered by Chloe
the interval are closed
Answered by oobleck
sorry. I missed the extra x
Answered by Chloe
i wrote it exactly as it was presented and after checking 3 is not the max
Answered by Chloe
all good i cant say anything i am the one struggling with the problem
Answered by oobleck
g(x) = x√(9-x^2)
g' = (9-2x^2)/√(9-x^2)
local max at x = 3/√2
g(0) = 0
g(3) = 0
so abs max is 9/2
abs min = 0
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