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-2 | -5 | global max  (vertex)

Step-by-step 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:

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| f(x) has a global maximum at x = -2

the first one, all those points on the line just make a strait line.