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.

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.)单项选择题

类似问题

Let [math: f] be a continuous function defined on the domain [math: [0,2]][0,2]. If [math: f(0)=1] and [math: f(2)=3], then the equation [math: f(x)=0] has no solution.

Consider the function 𝑔 ( 𝑥 ) = 1 𝑥 on the interval [ − 1 , 1 ] . We know that 𝑔 ( − 1 ) = − 1 and 𝑔 ( 1 ) = 1 . Which of the following statements is correct?

Consider the function g(x)= 1 x on the interval [−1,1]. We know that g(−1)=−1 and g(1)=1. Which of the following statements is correct?

Consider the function on the interval . We know that and . Which of the following statements is correct?

Name the Supreme Court decision that removed financial constraints on PACs.

Political scientists Joshua Kalla and David Broockman find that senior policy makers made themselves available to _____ between three and four times more often than they did to constituents.

Name the political scientist who argues that parties are designed to serve the needs of ambitious politicians (namely candidates and officeholders).

更多留学生实用工具

智能学习助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单