Question

question 6 (1 point) suppose f(x) is continuous on 1,2, f(1)=5, and f(2)=9. by the intermediate value theorem, we can conclude (choose all that apply): there exists a c between 1 and 2 such that f(c)=0. there exists a c between 1 and 2 such that f(c)=10. there exists a c between 1 and 2 such that f(c)=8. f(1.5)=7 for any y between 5 and 9, there exists a c between 1 and 2 such that f(c)=y. view hint for question 6

Explanation:

Step1: Recall Intermediate Value Theorem

If \(y = f(x)\) is continuous on \([a,b]\), and \(k\) is a number between \(f(a)\) and \(f(b)\), then there exists at least one number \(c\in(a,b)\) such that \(f(c)=k\). Here \(a = 1\), \(b = 2\), \(f(1)=5\) and \(f(2)=9\).

Step2: Analyze each option

• Option 1: Since \(0\) is not between \(5\) and \(9\), we cannot conclude there exists a \(c\in(1,2)\) such that \(f(c)=0\).
• Option 2: Since \(10\) is not between \(5\) and \(9\), we cannot conclude there exists a \(c\in(1,2)\) such that \(f(c)=10\).
• Option 3: Since \(8\) is between \(5\) and \(9\), by the Intermediate - Value Theorem, there exists a \(c\in(1,2)\) such that \(f(c)=8\).
• Option 4: Just because \(f(x)\) is continuous on \([1,2]\) and \(f(1) = 5\), \(f(2)=9\), we cannot say \(f(1.5)=7\). There is no information about the linearity or the exact form of \(f(x)\).
• Option 5: By the Intermediate - Value Theorem, for any \(y\) between \(5\) and \(9\), there exists a \(c\in(1,2)\) such that \(f(c)=y\).

Answer:

There exists a \(c\) between \(1\) and \(2\) such that \(f(c)=8\); For any \(y\) between \(5\) and \(9\), there exists a \(c\) between \(1\) and \(2\) such that \(f(c)=y\)