
Problem 11.1. Consider the function $f(x)=\sqrt{x}+\sqrt{x-4}-$ $\sqrt{x-1}-\sqrt{x-3}, x \geq 4$.

a.) Find $\lim _{x \rightarrow \infty} f(x)$.

b.) Prove that $f(x)$ is an increasing function.

c.) Find the number of real roots of the equation $f(x)=a \sqrt{\frac{x-3}{x}}$, where $a$ is a real parameter.


Solution: a.) By grouping the first and third radicals and the second and fourth radicals and rationalising we get that when $x>4$, $f(x)=\frac{1}{\sqrt{x}+\sqrt{x-1}}-\frac{1}{\sqrt{x-4}+\sqrt{x-3}}$. Therefore $\lim _{x \rightarrow \infty} f(x)=0$.

b.) When $x>4$,

$$
\begin{aligned}
f^{\prime}(x)= & \frac{1}{2 \sqrt{x}}-\frac{1}{2 \sqrt{x-1}}-\frac{1}{2 \sqrt{x-3}}+\frac{1}{2 \sqrt{x-4}} \\
= & \frac{1}{2 \sqrt{x-3} \sqrt{x-4}(\sqrt{x-3}+\sqrt{x-4})}+ \\
& -\frac{1}{2 \sqrt{x} \sqrt{x-1}(\sqrt{x}+\sqrt{x-1})}>0 .
\end{aligned}
$$

Therefore $f(x)$ is an increasing function if $x \geq 4$.

c.) It follows from a.) and b.) that $f(x)<0$ when $x \geq 4$, i. e., the equation could have a solution only if $a<0$. Let $a<0$. The function $g(x)=a \sqrt{\frac{x-3}{x}}=a \sqrt{1-\frac{3}{x}}$ is decreasing and continuous and $\lim _{x \rightarrow \infty} g(x)=a<0$. Since $f(x)$ is an increasing and continuous function and $\lim _{x \rightarrow \infty} f(x)=0$, in accordance with the Bolzano-Weierstraß Theorem the equation $f(x)=g(x)$ has a solution (and it is a unique one) exactly when $f(4) \leq g(4)$, i. e., if $2(1-\sqrt{3}) \leq a<0$.
