Question

the table gives selected values of the continuous function f.

x	-3	0	5	17

can we use the intermediate value theorem to say the equation f(x) = 150 has a solution where 0 ≤ x ≤ 5?

choose 1 answer:

a) no, since we dont know if the function is continuous on that interval.

b) no, since 150 is not between f(0) and f(5).

c) yes, both conditions for using the intermediate value theorem have been met.

Explanation:

The Intermediate Value Theorem (IVT) states that if a function \( f \) is continuous on a closed interval \([a, b]\), and \( k \) is a number between \( f(a) \) and \( f(b) \), then there exists at least one \( c \) in \([a, b]\) such that \( f(c)=k \).

First, check the continuity: the problem states \( f \) is continuous. For the interval \([0, 5]\), we find \( f(0) = 230 \) and \( f(5)=175 \). Now, check if \( 150 \) is between \( f(0) \) and \( f(5) \). We have \( 175<150<230 \)? No, \( 150 \) is between \( 175 \) and \( 230 \)? Wait, \( 175 < 230 \), and \( 150 \) is less than \( 175 \)? Wait, no: \( f(0) = 230 \), \( f(5)=175 \). So the values of \( f \) on \([0,5]\) go from \( 230 \) to \( 175 \). So the range of \( f \) on \([0,5]\) is between \( 175 \) and \( 230 \) (since it's continuous, it takes all values between \( 175 \) and \( 230 \)). Wait, \( 150 \) is less than \( 175 \), so \( 150 \) is not between \( f(0) = 230 \) and \( f(5)=175 \) (because \( 150 < 175 \), and \( 175 < 230 \), so the order is \( 150 < 175 < 230 \), so \( 150 \) is not between \( 175 \) and \( 230 \)). Wait, no: the IVT requires \( k \) to be between \( f(a) \) and \( f(b) \), regardless of which is larger. So if \( f(a) > f(b) \), then \( k \) must be between \( f(b) \) and \( f(a) \). Here, \( f(0) = 230 \), \( f(5)=175 \), so the interval of \( f \)-values is \([175, 230]\) (since it's continuous, it decreases from \( 230 \) to \( 175 \)). Now, \( 150 \) is less than \( 175 \), so \( 150 \) is not between \( 175 \) and \( 230 \). Wait, no: \( 175 < 230 \), so the values between \( f(5) \) and \( f(0) \) are from \( 175 \) to \( 230 \). \( 150 \) is less than \( 175 \), so it's not in that interval. Wait, but wait: \( f(0) = 230 \), \( f(5)=175 \). So the function goes from \( 230 \) to \( 175 \) on \([0,5]\). So the values of \( f(x) \) on \([0,5]\) are between \( 175 \) and \( 230 \) (since it's continuous, it must take all values between \( 175 \) and \( 230 \)). \( 150 \) is less than \( 175 \), so it's not between \( 175 \) and \( 230 \). Therefore, \( 150 \) is not between \( f(0) \) and \( f(5) \), so IVT does not guarantee a solution here.

Wait, but let's re - check: \( f(0) = 230 \), \( f(5)=175 \). So \( 175\leq f(x)\leq230 \) for \( x\in[0,5] \) (because it's continuous and decreasing? Wait, not necessarily decreasing, just continuous. But even if it's not decreasing, by IVT, if it's continuous on \([0,5]\), then it takes all values between \( f(0) \) and \( f(5) \). So if \( f(0)=230 \) and \( f(5)=175 \), then the values of \( f(x) \) on \([0,5]\) are between \( 175 \) and \( 230 \). \( 150 \) is less than \( 175 \), so it's not in that interval. Therefore, \( 150 \) is not between \( f(0) \) and \( f(5) \), so we can't use IVT to say \( f(x)=150 \) has a solution in \([0,5]\).

Option A is wrong because the problem states \( f \) is continuous. Option C is wrong because \( 150 \) is not between \( f(0) \) and \( f(5) \). So the correct answer is B.

Answer:

B. No, since 150 is not between \( f(0) \) and \( f(5) \).