Let $f$ be a twice differentiable function with $f(-1) = 1 = f(1)$ and $f(0) = 0$. Then the minimum number of roots of the equation
$$f(x)\bigl(f'(x)\bigr)^3 + \bigl(f(x)\bigr)^2 f'(x)f''(x) - f(x)f'(x) = x\bigl(f'(x)\bigr)^2 + xf(x)f''(x) - x$$
is/are
Step-by-Step Solution
Key Concept: Factor the equation as $[f(x)f'(x) - x][(f'(x))^2 + f(x)f''(x) - 1] = 0$. Let $g(x) = f^2(x) - x^2$. Since $g(-1) = g(0) = g(1) = 0$, $g$ has 3 roots, so $g'(x) = 0$ has at least 2 roots by Rolle's theorem (and $g'(0)=0$ is already one). The second factor $\frac{d}{dx}[f(x)f'(x) - x] = 0$ has roots from both cases.
Factor: $(f(x)f'(x) - x)((f'(x))^2 + f(x)f''(x) - 1) = 0$. Let $g(x) = f^2(x) - x^2$. $g$ has roots at $x = -1, 0, 1$ (3 roots). By Rolle, $g'(x) = 0$ has at least 2 roots in $(-1,0)$ and $(0,1)$. Note $g'(0) = 2f(0)f'(0) = 0$ is also a root. Analyzing both factors together gives a minimum of $\mathbf{5}$ roots.
Correct Answer: 5