Question

question 10 of 10 - / 4 view policies current attempt in progress a function f has f(5)=22, f(5)=2 and f(x)<0, for x≥5. which of the following are possible values for f(7)? 24 26 28

Explanation:

Step1: Use the Mean - Value Theorem

By the Mean - Value Theorem, if \(y = f(x)\) is continuous on the closed interval \([a,b]\) and differentiable on the open interval \((a,b)\), then \(f(b)-f(a)=f^{\prime}(c)(b - a)\) for some \(c\in(a,b)\). Here, \(a = 5\), \(b = 7\), so \(f(7)-f(5)=f^{\prime}(c)(7 - 5)=2f^{\prime}(c)\) for some \(c\in(5,7)\).

Step2: Analyze the sign of the second - derivative

Since \(f^{\prime\prime}(x)<0\) for \(x\geq5\), the function \(y = f^{\prime}(x)\) is a decreasing function for \(x\geq5\). We know that \(f^{\prime}(5) = 2\), so \(f^{\prime}(c)<f^{\prime}(5)=2\) for \(c\in(5,7)\).

Step3: Find the range of \(f(7)\)

From \(f(7)-f(5)=2f^{\prime}(c)\), and \(f(5) = 22\), we have \(f(7)=22 + 2f^{\prime}(c)\). Since \(f^{\prime}(c)<2\), then \(f(7)=22+2f^{\prime}(c)<22 + 2\times2=26\).

Answer:

24