[GRE 1268, #13a] Suppose \(f\) is a differentiable function such that \(f(1)=3\) and \(x\leq f'(x)\leq x^2\) for all \(x\) in the interval \(0\leq x \leq 3\). Find the smallest closed interval that must contain \(f(2)\).

Solution This problem relies on two theorems:

1. The Fundamental Theorem of Calculus (Part II): for any function \(f\) defined and differentiable on some interval \([c,d]\), and any numbers \(a,b\) such that \(c\lt a\leq b \lt d\), we have \[\int_a^b f'(x) dx = f(b)-f(a).\]
2. If \(f\) and \(g\) are continuous functions and \(f(x) \leq g(x) \) for all \(x\) in \([a,b]\), then \[\int_a^b f(x)dx \leq \int_a^b g(x) dx.\]

Pay attention to the hypotheses of the FTC: the function needs to be defined and differentiable throughout the interval \([a,b]\). A typical example of a situation when this theorem is abused is when \(f(x)=1/x\) and (for example) \(a=-1\), \(b=1\). The function is surely defined and differentiable at \(1\) and \(-1\). But if we apply the theorem without checking the hypotheses we find \[\int_{-1}^1 f'(x) dx = f(1)-f(-1) = 1-(-1)=2.\] This cannot make any sense because \(f'(x)=-1/x^2\) is always negative and so the area between its graph and the \(x\)-axis cannot be a positive number. Of course the problem is that \(f(x)\) is not defined at \(x=0\), so the hypotheses of the theorem are not satisfied.

In the given problem we are told that \(x\leq f'(x)\leq x^2\) for all \(x\) in the interval \(0\leq x \leq 3\). So we can apply the FTC to any pair of numbers in the interval \([0,3]\). Since we are given the value \(f(1)\) and we are asked a question about the value \(f(2)\), a reasonable choice is \(a=1\), \(b=2\). Applying the FTC and using the given value \(f(1)=3\), we find \[\int_1^2 f'(x) dx =f(2)-f(1) =f(2)-3.\] We now use the second theorem: since \(x\leq f'(x)\leq x^2\) we must have \[\int_1^2 xdx \leq \int_1^2 f'(x)dx \leq \int_1^2 x^2 dx.\] The integral on the left is easily found to be \(3/2\) and the one on the right \(7/3\). So substitution the value for the integral in the middle we just found, we get \[\frac{3}{2} \leq f(2)-3 \leq \frac{7}{3}.\] Adding \(3\) to all three expressions, \[\frac{9}{2} \leq f(2) \leq \frac{16}{3}.\] So we conclude that \(f(2)\) must be in the interval \([9/2,16/3]\).