Discussion
Let \(f(x)\) and \(g(x)\) be twice differentiable functions defined on [0,2] such that \(f''(x) - g''(x) = 0, f'(1) = 4, g'(1) = 2, f(2) = 9, g(2) = 3\). At \(x = 3/2, f(x) - g(x) =\)
Answer: C
\(f(x) - g(x) = 5\) at \(x = \frac{3}{2}\).
Let \(h = f - g\). Then \(h'' = 0\), so \(h\) is linear: \(h(x) = px + q\).
\[ \begin{aligned} h'(1) &= 4 - 2 = 2 = p \\ h(2) &= 9 - 3 = 6 = 2p + q, \quad q = 2 \\ h\left(\tfrac{3}{2}\right) &= 2 \times \tfrac{3}{2} + 2 = 5 \end{aligned} \]
6 is the value at \(x = 2\) (not offered) and 2 is the slope; neither is the value at \(\frac{3}{2}\).
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