Asked by Havon
Which statement is true about the given function?
f(x) < 0 over the intervals (–infinity, –2.7) and (0.8, infinity).
f(x) < 0 over the intervals (–infinity, –2.7) and
(–1, 0.8).
f(x) > 0 over the intervals (–infinity, –2.7) and
(–1, 0.8).
f(x) > 0 over the intervals (–infinity, –1) and
(0.8, infinity).
All Answers 1
Answered by
GPT-5 mini
AI
I can’t tell from what you posted — you omitted the function or its graph. To decide which statement is true you must know the sign of f on each interval determined by the zeros x = −2.7, −1, and 0.8. Here’s how to determine it:
1. Mark the zeros −∞ … −2.7 … −1 … 0.8 … ∞ to get four test intervals.
2. Pick one test x in each interval (for example x = −3, −2, 0, 1).
3. Evaluate f at each test x (or read the graph). The sign of f at the test point is the sign on that whole interval unless the zero has even multiplicity (where the sign does not change).
If you meant the simple cubic f(x) = (x + 2.7)(x + 1)(x − 0.8) (distinct linear factors with positive leading coefficient), testing gives
- f(−3) < 0 on (−∞, −2.7),
- f(−2) > 0 on (−2.7, −1),
- f(0) < 0 on (−1, 0.8),
- f(1) > 0 on (0.8, ∞).
So in that case f(x) < 0 on (−∞, −2.7) and (−1, 0.8) (the second choice).
If you paste the function or a picture of the graph I’ll test the intervals and give the correct option.
1. Mark the zeros −∞ … −2.7 … −1 … 0.8 … ∞ to get four test intervals.
2. Pick one test x in each interval (for example x = −3, −2, 0, 1).
3. Evaluate f at each test x (or read the graph). The sign of f at the test point is the sign on that whole interval unless the zero has even multiplicity (where the sign does not change).
If you meant the simple cubic f(x) = (x + 2.7)(x + 1)(x − 0.8) (distinct linear factors with positive leading coefficient), testing gives
- f(−3) < 0 on (−∞, −2.7),
- f(−2) > 0 on (−2.7, −1),
- f(0) < 0 on (−1, 0.8),
- f(1) > 0 on (0.8, ∞).
So in that case f(x) < 0 on (−∞, −2.7) and (−1, 0.8) (the second choice).
If you paste the function or a picture of the graph I’ll test the intervals and give the correct option.
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