Question

If f(x) is an odd continuous function in [-2, 2] and differentiable in (-2, 2), then:

• A. f(c) must be zero for at least one c in (-2, 0)
• B. f'(2) + f'(-2) = 0
• C. f(c) must be zero for at least one c in (0, 2)
• D. f'(c) - 1/2 * f(2) = 0 for some c in (-2, 0)

Answer 1

Answer: (D)

f'(c) - 1/2 * f(2) = 0 for some c in (-2, 0)

Answer 2

1. An odd continuous function has $f(0)=0$. The derivative of an odd differentiable function is an even function ($f'(-x) = f'(x)$).
2. Options 1 and 3 are false as counterexamples like $f(x)=x^3$ show that roots are not guaranteed in $(-2,0)$ or $(0,2)$.
3. Option 2 is false because $f'(x)$ being even implies $f'(2)=f'(-2)$, so $f'(2)+f'(-2)=0$ means $f'(2)=0$, which is not always true (e.g., $f(x)=x$ has $f'(2)=1$).
4. Option 4 is true by applying the Mean Value Theorem to $f(x)$ on the interval $[-2, 0]$. MVT guarantees a $c \in (-2, 0)$ such that $f'(c) = \frac{f(0)-f(-2)}{0-(-2)}$. Using $f(0)=0$ and $f(-2)=-f(2)$, we get $f'(c) = \frac{0-(-f(2))}{2} = \frac{f(2)}{2}$, which is $f'(c) - \frac{1}{2}f(2) = 0$.