Question

if ( f(4)=3 ) and ( f^{prime}(x) geq 3 ) for ( 4 leq x leq 9 ), how small can ( f(9) ) possibly be? ( f(9) geq )

Explanation:

Step1: Apply the Mean Value Theorem

The Mean Value Theorem states that if \(y = f(x)\) is continuous on \([a,b]=[4,9]\) and differentiable on \((a,b)=(4,9)\), then \(f^{\prime}(c)=\frac{f(b)-f(a)}{b - a}\) for some \(c\in(a,b)\). Here, \(a = 4\), \(b = 9\), so \(f^{\prime}(c)=\frac{f(9)-f(4)}{9 - 4}=\frac{f(9)-3}{5}\).

Step2: Use the condition \(f^{\prime}(x)\geq3\)

Since \(f^{\prime}(x)\geq3\) for \(4\leq x\leq9\), then \(f^{\prime}(c)\geq3\). Substituting \(f^{\prime}(c)=\frac{f(9)-3}{5}\) into \(f^{\prime}(c)\geq3\), we get \(\frac{f(9)-3}{5}\geq3\).

Step3: Solve the inequality for \(f(9)\)

Multiply both sides of the inequality \(\frac{f(9)-3}{5}\geq3\) by \(5\): \(f(9)-3\geq15\). Then add \(3\) to both sides: \(f(9)\geq15 + 3\).

Answer:

\(18\)