Question

if ( f(4)=9 ) and ( f^{prime}(x) geq 2 ) for ( 4 leq x leq 7 ), how small can ( f(7) ) possibly be? ( f(7) geq )

Explanation:

Step1: Apply the Mean Value Theorem

The Mean Value Theorem states that if \(y = f(x)\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then \(f(b)-f(a)=f^{\prime}(c)(b - a)\) for some \(c\in(a,b)\). Here, \(a = 4\), \(b = 7\), so \(f(7)-f(4)=f^{\prime}(c)(7 - 4)=3f^{\prime}(c)\), where \(4\lt c\lt7\).

Step2: Use the given condition \(f^{\prime}(x)\geq2\)

Since \(f^{\prime}(c)\geq2\) (because \(f^{\prime}(x)\geq2\) for \(4\leq x\leq7\)), then \(f(7)-f(4)\geq3\times2\).

Step3: Substitute \(f(4) = 9\)

We know \(f(4) = 9\). So \(f(7)-9\geq6\). Add 9 to both sides of the inequality: \(f(7)\geq9 + 6\).

Answer:

\(15\)