5. - / 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)