Suppose that
𝑓
(
𝑥
)
is a function that is a rational function and its domain is
(
−
∞
,
−
3
)
∪
(
−
3
,
0
)
∪
[
0
,
∞
)
. Also suppose that
𝑓
(
2
)
=
−
1
and
𝑓
(
4
)
=
1
. What may we conclude about
𝑓
(
𝑥
)
? (There is only one correct answer.)

Question

Suppose that 
𝑓
(
𝑥
)
 is a function that is a rational function and its domain is 
(
−
∞
,
−
3
)
∪
(
−
3
,
0
)
∪
[
0
,
∞
)
 . Also suppose that 
𝑓
(
2
)
=
−
1
 and 
𝑓
(
4
)
=
1
. What may we conclude about 
𝑓
(
𝑥
)
? (There is only one correct answer.)

BlackTom AI · Accepted Answer

Question restatement: We are told f is a rational function with domain (-∞, -3) ∪ (-3, 0) ∪ [0, ∞). It is given that f(2) = -1 and f(4) = 1. What may we conclude about f(x)? (There is only one correct answer.)

Option under consideration: 'We may conclude that f(c) = 0 for some c, by the Intermediate Value Theorem.'

Analysis of the option:
- Step 1: Identify a region where the function is continuous. Rational functions are continuous at all points in their domain. The interval [2, 4] lies entirely within the domain (since 2 and 4 are in [0, ∞) part of the domain and the function is defined there), so f is continuous on [2, 4].
- Step 2: Check endpoint values. We are given f(2) = -1 and f(4) = 1. The function takes opposite signs at the endpoints of the interval [2, 4].
- Step 3: Apply the Intermediate Value Theorem (IVT). The IVT says that if a function is continuous on [a, b] and takes values f(a) and f(b) with opposite signs, then there exists some c in (a, b) such that f(c) = 0. Here, a = 2 and b = 4, so there must exist some c in (2, 4) with f(c) = 0.
- Step 4: Address potential concerns about the domain? The IVT is applicable because the entire interval [2, 4] is within the function’s domain, so continuity on [2, 4] is guaranteed. There is no need to consider the domain gaps at -3 or 0 for this particular interval.

Therefore, the conclusion stated in the option is valid: there exists some c such that f(c) = 0, guaranteed by the Intermediate Value Theorem given the provided values and the continuity of f on the interval [2, 4].

Note on potential alternatives (implicit): If an option claimed something about zeros outside a continuous interval where f is defined, or relied on a non-continuous region, it would be questionable. However, in this case, the stated conclusion is precisely what IVT guarantees under the given conditions.

MATH-112-301-001 Unproctored Midcourse 1 Practice Exam 2

Suppose that 𝑓 ( 𝑥 ) is a function that is a rational function and its domain is ( − ∞ , − 3 ) ∪ ( − 3 , 0 ) ∪ [ 0 , ∞ ) . Also suppose that 𝑓 ( 2 ) = − 1 and 𝑓 ( 4 ) = 1 . What may we conclude about 𝑓 ( 𝑥 ) ? (There is only one correct answer.)

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