How do you calculate the minimum values of $f(x)=\frac{x^2+7/4}{\sqrt{x^2+3/4}}$?

Question

How do you calculate the minimum values of $f(x)=\frac{x^2+7/4}{\sqrt{x^2+3/4}}$?

I have no idea about this, but I know the answer is $x = \pm \frac{1}{2}$.

Answer 1

Let $y = \sqrt{x^2+\frac{3}{4}}$. Then $$f(x) = g(y) = y+\frac{1}{y}.$$ By the AM-GM inequality, $y+1/y \ge 2$, so the minimum of $f(x)$ is $2$. The minimum value of $f(x)$ is achieved when $y=1 \implies x^2+3/4=1 \implies x^2 =1/4 \implies x = \pm 1/2$.

Answer 2

The minimums of $f$ coincide with the minimums of $\ln f$, so

\begin{align} \frac{d}{dx}\ln f(x) &= \frac{d}{dx}\ln\left(\frac{x^2+7/4}{\sqrt{x^2+3/4}}\right) = \frac{d}{dx}\ln(x^2+7/4) - \frac{1}{2} \frac{d}{dx}\ln(x^2+3/4)\\ &= \frac{2x}{x^2+7/4} - \frac{1}{2}\cdot\frac{2x}{x^2+3/4} = \frac{x^3 - \tfrac14x}{(x^2+7/4)(x^2+3/4)}\\ &= \frac{x(x-1/2)(x+1/2)}{(x^2+7/4)(x^2+3/4)} \end{align}

So there are critical points at $x=0, \pm 1/2$. At the latter two are equal due to even-ness of $f$, i.e., $f(-x)=f(x)$. Computation gives $f(-1/2)=f(1/2) < f(0)$, so $x=0$ is not a minimum and given that $f(x)\to\infty$ as $x\to \pm \infty$, we conclude that $f$ has global minima at $x=\pm1/2$ by continuity of $f$ on $\mathbb{R}$.