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.
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:
8 answers
max as 3 is not correct
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.
√(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.
the interval are closed
sorry. I missed the extra x
i wrote it exactly as it was presented and after checking 3 is not the max
all good i cant say anything i am the one struggling with the problem
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
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
1 answer