Discussion
Assume that \(f:[0,1] \to \mathbb{R}\) is continuous on [0,1] and differentiable on (0,1) such that \(f(x + h) = f(x) + hf'(x + \theta h)\) for some \(0 \lt \theta \lt 1\). If \(f(x) = x^{2}(1 + x)\), and \(\theta\) is expressed in terms of \(x\) and \(h\), then the value of \(\lim_{h \to 0} \theta(x,h)\) is
Answer: B
The limit is \(\frac{1}{2}\).
Expand both sides of \(f(x + h) - f(x) = hf'(x + \theta h)\) in powers of h, using Taylor's theorem. For the cubic \(f = x^{2} + x^{3}\), \(f''' = 6\) is constant:
\[ \begin{aligned} \text{LHS} &= hf' + \tfrac{h^{2}}{2}f'' + \tfrac{h^{3}}{6}f''' \\ \text{RHS} &= hf' + \theta h^{2}f'' + \tfrac{\theta^{2}h^{3}}{2}f''' \end{aligned} \]
Cancelling \(hf'\) and dividing by \(h^{2}\): \(\tfrac{1}{2}f'' + \tfrac{h}{6}f''' = \theta f'' + \tfrac{\theta^{2}h}{2}f'''\). As \(h \to 0\) this leaves \(\theta f'' = \tfrac{1}{2}f''\), so \(\theta \to \tfrac{1}{2}\) wherever \(f''(x) = 2 + 6x \neq 0\).
The general result is that \(\theta \to 1/2\) for any twice-differentiable f with \(f'' \neq 0\), so 1/3, 1/4 and 4/5 are ruled out.
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