Question

Find the local maximum and minimum values of the function and the value of x at which the following occurs. state the following answer correct to two decimal places. u(x)=6 *sqrt(6-c)

a) Local max: x = 0.00, Local min: x = 6.00
b) Local max: x = 2.00, Local min: x = 4.00
c) Local max: x = 1.50, Local min: x = 4.50
d) Local max: x = 3.00, Local min: x = 5.00

Answer 1

To find the local maximum and minimum values of the function u(x) = 6sqrt(6-c), find the critical points by taking the derivative. Check the endpoints of the interval (0, 6) as well. The local maximum is at x = 6 and the local minimum is at x = 0.

Step-by-step explanation:

To find the local maximum and minimum values of the function u(x) = 6sqrt(6-c), we first need to find the critical points by taking the derivative. The derivative is u'(x) = 6/(2sqrt(6-c)). Setting u'(x) equal to zero and solving for x gives us x = 0. Therefore, x = 0 is a critical point. We need to also check the endpoints of the interval (0, 6) since they could potentially be the maximum or minimum values. Plugging in x = 0 and x = 6 into u(x), we get u(0) = 6sqrt(6) and u(6) = 6sqrt(0), which simplifies to 6(0) = 0. This gives us a local minimum at x = 0 and a local maximum at x = 6.