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)…

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)