Answer:
-2 | -5 | global max (vertex)
Explanation:
Find and classify the global extrema of the following function:
f(x) = -abs(x + 2)/3 - 5
Find the critical points of f(x):
Compute the critical points of -abs(x + 2)/3 - 5
To find all critical points, first compute f'(x):
d/( dx)(-abs(x + 2)/3 - 5) = -abs(x + 2)/(3 (x + 2)):
f'(x) = -abs(x + 2)/(3 (x + 2))
f'(x) is never zero on the real line:
-abs(x + 2)/(3 (x + 2)) is never zero
f'(x) does not exist at x = -2:
x = -2
The only critical point of -abs(x + 2)/3 - 5 is at x = -2:
x = -2
The domain of -abs(x + 2)/3 - 5 is R:
The endpoints of R are x = -∞ and ∞
Evaluate -abs(x + 2)/3 - 5 at x = -∞, -2 and ∞:
The open endpoints of the domain are marked in gray
x | f(x)
-∞ | -∞
-2 | -5
∞ | -∞
The largest value corresponds to a global maximum, and the smallest value corresponds to a global minimum:
The open endpoints of the domain are marked in gray
x | f(x) | extrema type
-∞ | -∞ | global min
-2 | -5 | global max
∞ | -∞ | global min
Remove the points x = -∞ and ∞ from the table
These cannot be global extrema, as the value of f(x) here is never achieved:
x | f(x) | extrema type
-2 | -5 | global max
f(x) = -abs(x + 2)/3 - 5 has one global maximum:
Answer: |
| f(x) has a global maximum at x = -2