Let f : R rightarrow R be a continuous odd fu | vedclass

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(B) Since $f$ is an odd function,$f(-t) = -f(t)$.$F(1) = \int_{-1}^1 f(t) dt = 0$ because the integral of an odd function over a symmetric interval $[

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$7$

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(B) Since $f$ is an odd function,$f(-t) = -f(t)$.$F(1) = \int_{-1}^1 f(t) dt = 0$ because the integral of an odd function over a symmetric interval $[-a, a]$ is $0$.$G(1) = \int_{-1}^1 t|f(f(t))| dt = 0$ because the integrand $h(t) = t|f(f(t))|$ is an odd function $(h(-t) = -t|f(f(-t))| = -t|f(-f(t))| = -t|f(f(t))| = -h(t))$.Using $L'H\hat{o}pital's$ rule for the limit $\lim_{x \rightarrow 1} \frac{F(x)}{G(x)}$:$\lim_{x \rightarrow 1} \frac{F'(x)}{G'(x)} = \lim_{x \rightarrow 1} \frac{f(x)}{x|f(f(x))|} = \frac{f(1)}{1 \cdot |f(f(1))|} = \frac{1/2}{|f(1/2)|} = \frac{1}{14}$.Therefore,$|f(1/2)| = 7$. Since $f$ is a continuous odd function vanishing at only one point (which must be $x=0$) and $f(1) = 1/2 > 0$,$f(x)$ must be positive for $x > 0$. Thus,$f(1/2) = 7$.

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