The function f, of, xf(x) is continuous on its domain of open square bracket, minus, 9, comma, 9, close square bracket[−9,9] and is plotted below such that the portion of the graph on the interval left parenthesis, minus, 4, comma, 3, right parenthesis(−4,3) is hidden from view. Given that f, of, minus, 4, equals, minus, 5f(−4)=−5 and f, of, 3, equals, minus, 2, commaf(3)=−2, determine what conclusions can be drawn based on the Intermediate Value Theorem on the interval left parenthesis, minus, 4, comma, 3, right parenthesis, .(−4,3).
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Answer
Attempt 1 out of 2

Since the function f, of, xf(x) is 
 on its domain, the Intermediate Value Theorem 
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We are given that the function \( f(x) \) is continuous on the interval \([-9, 9]\), and we are interested in applying the **Intermediate Value Theorem** (IVT) on the interval \( (-4, 3) \).

### Restating the problem:
- \( f(-4) = -5 \)
- \( f(3) = -2 \)
- The portion of the graph between \( x = -4 \) and \( x = 3 \) is hidden.
  
The **Intermediate Value Theorem** states that if a function \( f \) is continuous on a closed interval \([a, b]\) and \( f(a) \neq f(b) \), then for any value \( N \) between \( f(a) \) and \( f(b) \), there exists some \( c \in (a, b) \) such that \( f(c) = N \).

### Applying IVT:
- The function \( f(x) \) is continuous on the interval \((-4, 3)\), which implies that for any value \( N \) between \( f(-4) = -5 \) and \( f(3) = -2 \), there must exist some \( c \in (-4, 3) \) such that \( f(c) = N \).
- Since \( f(-4) = -5 \) and \( f(3) = -2 \), the IVT guarantees that the function \( f(x) \) must take **every value between -5 and -2** somewhere on the interval \( (-4, 3) \).

### Conclusion:
From the Intermediate Value Theorem, we can conclude that there exists at least one value \( c \in (-4, 3) \) for each of the following:
1. \( f(c) = -4 \)
2. \( f(c) = -3 \)
3. \( f(c) = -2.5 \) (or any other value between -5 and -2)

In summary, the IVT guarantees that the function takes on every value between \( -5 \) and \( -2 \) on the hidden interval \( (-4, 3) \), even though the exact behavior of the function in this interval is not visible.

Would you like further details or have any questions?

Here are 5 related questions:
1. What conditions are required for the Intermediate Value Theorem to apply?
2. Can the Intermediate Value Theorem tell us how many times \( f(x) \) takes a particular value?
3. How does continuity of \( f(x) \) affect its graph on the interval \( (-4, 3) \)?
4. What if the function were not continuous on \( (-4, 3) \)? Would the IVT still apply?
5. How does the IVT relate to finding roots of an equation?

**Tip:** When applying the IVT, always check that the function is continuous on the given interval.

The function f(x) is continuous on its domain of [-9, 9] and is plotted such that the portion of the graph on the interval (-4, 3) is hidden from view. Given that f(-4) = -5 and f(3) = -2, determine what conclusions can be drawn based on the Intermediate Value Theorem on the interval (-4, 3).

Learn how the Intermediate Value Theorem applies to a continuous function over the interval (-4, 3). The function is continuous and takes on every value between -5 and -2 within this hidden range. Understand key concepts in continuity and function behavior.

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