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aalyn [17]
3 years ago
10

Can someone help me with this problem

Mathematics
2 answers:
expeople1 [14]3 years ago
7 0
Increasing
Increasing
Decreasing
t=3
t=14
V=4
UNO [17]3 years ago
7 0
<h3>Answers:</h3>
  • A) The graph of the function is <u>increasing</u> over the interval 0 < t < 3
  • B) The graph of the function is <u>increasing</u> over the interval 14 < t < 18
  • C) The graph of the function is <u>decreasing</u> over the interval 3 < t < 14
  • D) The graph has a max value of V = 7 at <u>t = 3</u>
  • E) The local min (V = 1) occurs when <u>t = 14</u>
  • F) I'm not sure what your teacher is trying to say here. There isn't a local max or local min when t = 18. Seems like a typo. If you ignore the "local max" part and you're only wanting the value of V, then V = 4 here.

========================================================

Explanation:

The graph goes uphill as you read from left to right over the interval 0 < t < 3. This means t is between 0 and 3. So the graph is increasing over this interval. The same applies for the interval 14 < t < 18 as well.

The graph goes downhill we we are between t = 3 and t = 14, so that's why the graph is decreasing on the interval 3 < t < 14.

The local min is the lowest point of the graph. For sine functions like this, they occur infinitely many often. But we're restricting to just one cycle (so to speak) and we only have one local min at (t, V) = (14, 1). So that explains the answer to part E. The answer to part D is similar but we're looking at the highest point this time at (3,7)

As for part F, it seems like your teacher made a typo because the graph is neither at the highest nor the lowest point when t = 18. The graph is right in the middle instead.

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Double-check to ensure that you have copied this problem down completely and accurately.

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Strike441 [17]
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because

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