Question

let $f(x) = x \cdot 3^x$.
can we use the intermediate value theorem to say the equation $f(x) = 100$ has a solution where $2 \leq x \leq 4$
choose 1 answer:
a no, since the function is not continuous on that interval.
b no, since 100 is not between $f(2)$ and $f(4)$.
c yes, both conditions for using the intermediate value theorem have been met.

Explanation:

Step1: Recall Intermediate Value Theorem (IVT) Conditions

The Intermediate Value Theorem states that if a function \( f(x) \) is continuous on a closed interval \([a, b]\), and \( k \) is a number between \( f(a) \) and \( f(b) \) (i.e., \( \min(f(a), f(b)) < k < \max(f(a), f(b)) \)), then there exists at least one \( c \) in \((a, b)\) such that \( f(c) = k \). First, we need to check the continuity of \( f(x) = x \cdot 3^x \) on \([2, 4]\). The function \( y = x \) is a polynomial (continuous everywhere), and \( y = 3^x \) is an exponential function (continuous everywhere). The product of two continuous functions is continuous, so \( f(x) = x \cdot 3^x \) is continuous on \([2, 4]\). So option A is incorrect.

Step2: Calculate \( f(2) \) and \( f(4) \)

- Calculate \( f(2) \): Substitute \( x = 2 \) into \( f(x) \). \( f(2) = 2 \cdot 3^2 = 2 \cdot 9 = 18 \).
- Calculate \( f(4) \): Substitute \( x = 4 \) into \( f(x) \). \( f(4) = 4 \cdot 3^4 = 4 \cdot 81 = 324 \).

Step3: Check if 100 is between \( f(2) \) and \( f(4) \)

We have \( f(2) = 18 \) and \( f(4) = 324 \). Since \( 18 < 100 < 324 \), 100 is between \( f(2) \) and \( f(4) \). Wait, but let's re - evaluate. Wait, no, wait: Wait, the function is increasing? Let's check the derivative to see if it's increasing. The derivative \( f^\prime(x)=3^x + x\cdot3^x\ln(3)\). For \( x\in[2,4]\), \( 3^x>0\), \( x > 0\), \( \ln(3)>0\), so \( f^\prime(x)>0\), so \( f(x) \) is increasing on \([2,4]\). So \( f(2)=18\), \( f(4)=324\), and 100 is between 18 and 324. But the options: Wait, the options are A, B, C. Wait, maybe I made a mistake. Wait, no, let's re - check the calculations. \( 3^2 = 9 \), \( 2\times9 = 18 \). \( 3^4=81 \), \( 4\times81 = 324 \). So 100 is between 18 and 324. But the option B says "No, since 100 is not between \( f(2) \) and \( f(4) \)". But 100 is between 18 and 324. Wait, maybe the original problem's options are mis - read? Wait, no, the user provided the options. Wait, maybe I miscalculated \( f(4) \). \( 3^4 = 81 \), \( 4\times81=324 \), correct. \( f(2)=18 \), correct. So 100 is between 18 and 324. Then the conditions for IVT are met: continuous on \([2,4]\) (since product of continuous functions is continuous), and 100 is between \( f(2) \) and \( f(4) \). So the answer should be C? But wait, the options: Wait, the user's options: A: No, not continuous. B: No, 100 not between. C: Yes, both conditions met. But according to our calculation, 100 is between 18 and 324, and the function is continuous. So the correct answer should be C. But wait, maybe I made a mistake in the derivative? Let's check \( f(2)=2\times3^2 = 18 \), \( f(4)=4\times3^4 = 4\times81 = 324 \). 18 < 100 < 324, so 100 is between them. And the function is continuous on \([2,4]\) (as product of continuous functions: polynomial and exponential are continuous, so their product is continuous). So both conditions for IVT are met, so the answer is C. But wait, the options: Let me re - check the problem statement. The problem is "Can we use the intermediate value theorem to say the equation \( f(x)=100 \) has a solution where \( 2\leq x\leq4 \)?" The IVT conditions are: 1. \( f \) is continuous on \([a,b]\); 2. \( k \) is between \( f(a) \) and \( f(b) \). Here, \( f(x)=x\cdot3^x \) is continuous on \([2,4]\) (as product of continuous functions), \( f(2)=18 \), \( f(4)=324 \), and 100 is between 18 and 324. So both conditions are met, so the answer is C. But wait, the options: Wait, the user's options: A: No, not continuous. B: No, 100 not between. C: Yes, both conditions met. So the correct answer is C. But wait, maybe I miscalculated \( f(4) \). Let's recalculate \( 3^4 \): \( 3\times3 = 9 \), \( 9\times3 = 27 \), \( 27\times3 = 81 \). Then \( 4\times81 = 324 \). Correct. \( f(2)=2\times9 = 18 \). Correct. So 100 is between 18 and 324, and the function is continuous. So the answer is C.

Wait, but the initial thought was wrong. Let's go back. The intermediate value theorem requires the function to be continuous on the closed interval and the value \( k \) to be between \( f(a) \) and \( f(b) \). Since \( f(x)=x\cdot3^x \) is continuous (product of continuous functions: \( y = x \) is continuous, \( y = 3^x \) is continuous, so their product is continuous) on \([2,4]\), and \( 18=f(2)<100<f(4)=324 \), so both conditions are met. So the answer is C.

Answer:

C. Yes, both conditions for using the intermediate value theorem have been met.