Consider the function f (x) = StartLayout Enlarged left-brace first row negative StartFraction x + 5 Over x + 3 EndFraction, x l
ess-than negative 2 second row x cubed + 6, x greater-than-or-equal-to negative 2 EndLayout.
Which statement describes whether the function is continuous at x = –2?
The function is continuous at x = –2 because f(–2) exists.
The function is continuous at x = –2 because Limit as x approaches negative 2 plus f(x) = f(–2).
The function is not continuous at x = –2 because Limit as x approaches negative 2 f(x) ≠ f(–2).
The function is not continuous at x = –2 because Limit as x approaches negative 2 f(x) does not exist.
Which statement describes whether the function is continuous at x = –2?
The function is continuous at x = –2 because f(–2) exists.
The function is continuous at x = –2 because Limit as x approaches negative 2 plus f(x) = f(–2).
The function is not continuous at x = –2 because Limit as x approaches negative 2 f(x) ≠ f(–2).
The function is not continuous at x = –2 because Limit as x approaches negative 2 f(x) does not exist.
1 answer:
You might be interested in
Answer:
The option "StartFraction 1 Over 3 Superscript 8" is correct
That is is correct answer
Therefore
Step-by-step explanation:
Given expression is ((2 Superscript negative 2 Baseline) (3 Superscript 4 Baseline)) Superscript negative 3 Baseline times ((2 Superscript negative 3 Baseline) (3 squared)) squared
The given expression can be written as
To find the simplified form of the given expression :
( using the property
)
( using the property
( combining the like powers )
( using the property
)
( using the property
)
Therefore
Therefore option "StartFraction 1 Over 3 Superscript 8" is correct
That is is correct answer