10. Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a smooth function such that $f^{\prime}(x)=f(1-x)$ for all $x$ and $f(0)=1$. Find $f(1)$.
Solution: $\sec 1+\tan 1$
Differentiating the given equation gives $f^{\prime \prime}(x)=-f(x)$. This has solutions of the form $A \cos (x)+B \sin (x)$. Since $f(0)=1, A=1$. Then $f^{\prime}(x)=B \cos (x)-\sin (x)$ and
$$
\begin{aligned}
f(1-x) & =\cos (1-x)+B \sin (1-x) \\
& =\cos 1 \cos x+\sin 1 \sin x+B \sin 1 \cos x-B \cos 1 \sin x \\
& =(\cos 1+B \sin 1) \cos x+(\sin 1-B \cos 1) \sin x
\end{aligned}
$$

Therefore, $B=\cos 1+B \sin 1$ and $-1=\sin 1-B \cos 1$, both of which yield as solutions
$$
B=\frac{\cos 1}{1-\sin 1}=\frac{1+\sin 1}{\cos 1}=\sec 1+\tan 1 .
$$