Final answer:
The function f(x) = -x² - x - 5 has a relative maximum at the point (1/2, -23/4), found by setting the derivative -2x - 1 equal to zero and confirming concavity with the second derivative.
Step-by-step explanation:
To find the x-values of all points where the function f(x) = −x² − x − 5 has any relative extrema, we first need to take the derivative of the function to locate potential points of extremum. The derivative of f(x) is f'(x) = -2x - 1. Setting the derivative equal to zero gives us the critical points, which are possible locations of relative extrema:
-2x - 1 = 0
x = -1 / -2
x = 1/2
To determine whether this critical point represents a maximum or minimum, we can either examine the second derivative or test the sign of the first derivative on intervals around x = 1/2. The second derivative of f(x) is f''(x) = -2, which is always negative, indicating that our function is concave down and thus the point x = 1/2 is a relative maximum.
To find the value of the relative maximum, we substitute x = 1/2 into the original function:
f(1/2) = −(1/2)² − 1/2 − 5
f(1/2) = −1/4 − 1/2 − 5
f(1/2) = −1/4 − 2/4 − 20/4
f(1/2) = −23/4
Therefore, the function has a relative maximum at the point (1/2, −23/4).