Question

1. - / 1 points if f(4) = 9 and f(x) ≥ 3 for 4 ≤ x ≤ 9, how small can f(9) possibly be? f(9) ≥

Explanation:

Step1: Apply the Mean - Value Theorem

By the Mean - Value Theorem, if \(y = f(x)\) is continuous on the closed interval \([a,b]=[4,9]\) and differentiable on the open interval \((a,b)=(4,9)\), then \(f(b)-f(a)=f^{\prime}(c)(b - a)\) for some \(c\in(a,b)\). Here, \(f(9)-f(4)=f^{\prime}(c)(9 - 4)\) where \(4\lt c\lt9\).

Step2: Substitute the known values

We know that \(f(4) = 9\) and \(f^{\prime}(x)\geq3\) for \(4\leq x\leq9\). So \(f(9)-9=f^{\prime}(c)\times5\). Since \(f^{\prime}(c)\geq3\), we have \(f(9)-9\geq3\times5\).

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

Adding 9 to both sides of the inequality \(f(9)-9\geq15\), we get \(f(9)\geq15 + 9\).

Answer:

\(f(9)\geq24\)