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:
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.
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.
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
To find the absolute maximum and minimum values of a function, we first need to find the critical points of the function within the given interval.
The function given is g(x) = x * sqrt(9 - x^2) defined on the interval [0, 3].
To find the critical points, we need to take the derivative of the function and set it equal to zero.
Taking the derivative of g(x), we use the product rule:
g'(x) = d/dx [x] * sqrt(9 - x^2) + x * d/dx [sqrt(9 - x^2)]
Using the chain rule, the derivative of sqrt(9 - x^2) is -(x / sqrt(9 - x^2)).
Simplifying:
g'(x) = sqrt(9 - x^2) + x * (-(x / sqrt(9 - x^2)))
g'(x) = sqrt(9 - x^2) - (x^2 / sqrt(9 - x^2))
Next, we set g'(x) equal to zero and solve for x:
sqrt(9 - x^2) - (x^2 / sqrt(9 - x^2)) = 0
Multiplying both sides by sqrt(9 - x^2), we get:
9 - x^2 - x^2 = 0
9 - 2x^2 = 0
Solving for x, we get:
2x^2 = 9
x^2 = 9/2
x = ± sqrt(9/2)
However, we need to check if these critical points are within the interval [0, 3].
For x = sqrt(9/2):
0 ≤ sqrt(9/2) ≤ 3
0 ≤ 3.0 ≤ 3
This lies within the interval.
For x = -sqrt(9/2):
0 ≤ -sqrt(9/2) ≤ 3
-3 ≤ 0.0 ≤ 3
This lies within the interval.
Now, we also need to consider the endpoints of the interval, x = 0 and x = 3.
So, the potential critical points are:
x = sqrt(9/2)
x = -sqrt(9/2)
x = 0
x = 3
To find the absolute maximum and minimum values of the function, we evaluate the function g(x) at these critical points as well as the endpoints.
g(0) = 0 * sqrt(9 - 0) = 0
g(3) = 3 * sqrt(9 - 3^2) = 3 * sqrt(9 - 9) = 3 * sqrt(0) = 3 * 0 = 0
g(sqrt(9/2)) = sqrt(9/2) * sqrt(9 - (sqrt(9/2))^2)
= sqrt(9/2) * sqrt(9 - 9/2)
= sqrt(9/2) * sqrt((18 - 9)/2)
= sqrt(9/2) * sqrt(9/2)
= 9/2
g(-sqrt(9/2)) = -sqrt(9/2) * sqrt(9 - (-sqrt(9/2))^2)
= -sqrt(9/2) * sqrt(9 - 9/2)
= -sqrt(9/2) * sqrt((18 - 9)/2)
= -sqrt(9/2) * sqrt(9/2)
= -9/2
Now, we need to compare these values to find the absolute maximum and minimum values.
The maximum value is the larger of these values, so the absolute maximum value is 9/2.
The minimum value is the smaller of these values, so the absolute minimum value is -9/2.
Therefore, the absolute maximum value is 9/2 and the absolute minimum value is -9/2 for the function g(x) = x * sqrt(9 - x^2) on the interval [0, 3].