Find the local maximum and minimum values of the function and the value of x at which each occurs. State each answer correct to two decimal places. (If an answer does not exist, enter DNE.)

U(x)=x√(5-x)

To find the local maximum and minimum values of the function U(x) = x√(5-x), we first need to find the critical points. Critical points occur where the derivative of the function is either zero or undefined.

1. Calculate the derivative of U(x) with respect to x:
U'(x) = √(5-x) - x/(2√(5-x))

2. Set U'(x) equal to zero to find the potential critical points:
√(5-x) - x/(2√(5-x)) = 0

3. To simplify the equation, multiply both sides by 2√(5-x) to eliminate the denominator:
2(5-x) - x = 0

4. Expand and combine like terms:
10 - 2x - x = 0
10 - 3x = 0
3x = 10
x = 10/3 ≈ 3.33

The critical point we found is x = 10/3 ≈ 3.33, but we need to check if it is a maximum or a minimum.

5. To determine whether it is a local maximum or minimum, we can use the second derivative test. Calculate the second derivative of U(x):

U''(x) = d^2(U(x))/dx^2 = (d(U'(x))/dx)

To find the second derivative, we differentiate U'(x) with respect to x:
U''(x) = (d(√(5-x) - x/(2√(5-x))))/dx

6. Simplify the second derivative:
U''(x) = (1/(2√(5-x))) - 1/(4(5-x)^(3/2))

Now evaluate U''(x) at x = 10/3 ≈ 3.33:

U''(3.33) = (1/(2√(5-3.33))) - 1/(4(5-3.33)^(3/2))

7. Evaluate and determine the value of U''(3.33):

U''(3.33) ≈ 0.177

Since U''(3.33) is positive, we can conclude that x = 10/3 ≈ 3.33 is a local minimum.

Therefore, the local minimum value of the function U(x) = x√(5-x) occurs at x ≈ 3.33 and is approximately equal to U(3.33).

To find the value of the local minimum, substitute x ≈ 3.33 into the function U(x):
U(3.33) = 3.33√(5-3.33)

8. Calculate the value of the local minimum:

U(3.33) ≈ 2.73

Therefore, the local minimum value of the function U(x) is approximately 2.73 and it occurs at x ≈ 3.33.

To find the local maximum, we need to check the endpoints of the interval.

9. Calculate the value of U(0):
U(0) = 0√(5-0) = 0

10. Calculate the value of U(5):
U(5) = 5√(5-5) = 0

Since both U(0) and U(5) are equal to 0, there are no local maximum values within the interval [0, 5].

In summary:
- The local minimum value of the function U(x) = x√(5-x) is approximately 2.73 and it occurs at x ≈ 3.33.
- There are no local maximum values within the interval [0, 5].