Discussion

Suppose that f is a function defined by :

\[ f(x) = \frac{x}{3 - 2x}, \quad 0 \lt x \lt 1 \]

If \(g(x) = f''(x) + f'(x) - [f'(x)]^{2}\), then g(1) is equal to :

Answer: C

\(g(1)\) equals 6.

Differentiate \(f(x) = \frac{x}{3 - 2x}\) by the quotient rule:

\[ \begin{aligned} f'(x) &= \frac{(3 - 2x) + 2x}{(3 - 2x)^{2}} = \frac{3}{(3 - 2x)^{2}} \\ f''(x) &= \frac{12}{(3 - 2x)^{3}} \end{aligned} \]

At \(x = 1\), \(3 - 2x = 1\), so \(f'(1) = 3\) and \(f''(1) = 12\).

\[ g(1) = 12 + 3 - 3^{2} = 6 \]

Forgetting to square \(f'(1)\) gives \(12 + 3 - 3 = 12\), not an option; dropping the \(f'(x)\) term gives 3. The value 9 is \([f'(1)]^{2}\) alone.

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