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let f be a function that is continuous on the closed interval 1,3 with …

Question

let f be a function that is continuous on the closed interval 1,3 with f(1)=10 and f(3)=18. which of the following statements must be true? a 10 ≤ f(2) ≤ 18 b f is increasing on the interval 1,3. c f(x)=17 has at least one solution in the interval 1,3. d f(x)=8 has at least one solution in the interval (1,3). e ∫₁³ f(x)dx > 20

Explanation:

Step1: Recall Intermediate - Value Theorem

The Intermediate - Value Theorem states that if a function \(y = f(x)\) 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 number \(c\) in the interval \((a,b)\) such that \(f(c)=k\). Here, \(a = 1\), \(b = 3\), \(f(1)=10\) and \(f(3)=18\).

Step2: Analyze option A

Just because \(f(1) = 10\) and \(f(3)=18\), we cannot say that \(10\leq f(2)\leq18\). The function is continuous, but it doesn't have to be monotonic. So, this statement is not necessarily true.

Step3: Analyze option B

We are only given the values of the function at the endpoints of the interval \([1,3]\). We have no information about the derivative of the function, so we cannot conclude that the function is increasing on the interval \([1,3]\).

Step4: Analyze option C

Since \(10<17<18\) and \(f(x)\) is continuous on \([1,3]\), by the Intermediate - Value Theorem, there exists at least one \(x\) in the interval \((1,3)\) (and thus in \([1,3]\)) such that \(f(x)=17\).

Step5: Analyze option D

Since \(8<10\), and \(f(x)\) is continuous on \([1,3]\), there is no guarantee that \(f(x)\) will take the value \(8\) in the interval \((1,3)\) as \(8\) is not between \(f(1) = 10\) and \(f(3)=18\).

Step6: Analyze option E

We know that \(f(x)\) is continuous on \([1,3]\), but we don't have enough information to determine the value of the definite - integral \(\int_{1}^{3}f(x)dx\). For example, the function could be very close to \(10\) for most of the interval \([1,3]\) and then jump to \(18\) near \(x = 3\), and \(\int_{1}^{3}f(x)dx\) may not be greater than \(20\).

Answer:

C. \(f(x)=17\) has at least one solution in the interval \([1,3]\)