Find all twice differentiable functions $f(x)$ such that $f^{\prime \prime}(x)=0, f(0)=19$, and $f(1)=99$.
Solution: Since $f^{\prime \prime}(x)=0$ we must have $f(x)=a x+b$ for some real numbers $a, b$. Thus $f(0)=$ $b=19$ and $f(1)=a+19=99$, so $a=80$. Therefore $f(x)=\mathbf{8 0} \mathbf{x}+\mathbf{1 9}$.