Find the absolute maximum value and the absolute minimum value, if any, of the function.
g(x)=(1/16)x^2-16sqrtx on [0,36]
To find the absolute maximum value and the absolute minimum value of the function g(x)=(1/16)x^2-16√x on the interval [0,36], we can follow these steps:
Step 1: Find the critical points of the function within the given interval.
To find the critical points, we need to find the values of x where the derivative of the function g'(x) equals zero or is undefined. Let's start with finding g'(x):
g(x) = (1/16)x^2 - 16√(x)
Taking the derivative with respect to x:
g'(x) = (1/16) * 2x - 16 * (1/2) * (1/√x)
= (1/8)x - 8/√x
Setting g'(x) equal to zero:
(1/8)x - 8/√x = 0
Simplifying the equation:
x/8 = 8/√x
Cross multiply:
x^2 = 64
Taking the square root of both sides:
|x| = 8
From this equation, we can determine two critical points: x = 8 and x = -8. However, since the interval is [0,36], we only consider the critical point x = 8.
Step 2: Calculate the values of the function at the endpoints and critical points within the given interval.
Evaluate the function g(x) at x = 0, x = 8, and x = 36:
g(0) = (1/16)(0)^2 - 16√(0) = 0
g(8) = (1/16)(8)^2 - 16√(8) ≈ -16
g(36) = (1/16)(36)^2 - 16√(36) ≈ -16
Step 3: Determine the absolute maximum and minimum values.
Comparing the values calculated in step 2, we can conclude the following:
The absolute maximum value of the function g(x) is 0, which occurs at x = 0.
The absolute minimum value of the function g(x) is approximately -16, which occurs at x = 8.
Therefore, the absolute maximum value and the absolute minimum value of the function g(x)=(1/16)x^2-16√x on the interval [0,36] are 0 and -16, respectively.