Showing that the graph of $\frac{\cos(\pi x/2)}{1-x^2}$ decreases over the interval $(0,2)$.

Question

How would one justify the behaviour of the graph of $$\frac{\cos(\frac{\pi}{2}x)}{1-x^2}$$ over the interval $(0,2)$?

More precisely, how can it be shown that the graph is decreasing over that interval?

Answer 1

The derivative of $f(x)$ takes the following form \begin{equation} f'(x) = \frac{g(x)}{(1-x^2)^2} \end{equation} \begin{equation} g(x) = -\frac{\pi}{2} \sin (\frac{\pi}{2}x)(1-x^2)+2x( \cos (\frac{\pi}{2}x)) \end{equation} PS: The function $f(x)$ is not defined at $1$

Since the denominator of $f'(x) > 0$ on $[0,1[ \cup ]1,2]$, we will study $g(x)$. Get it's derivative \begin{equation} g'(x) = -\frac{\pi^2}{4} \cos (\frac{\pi}{2}x)(1-x^2) + \pi x \sin (\frac{\pi}{2}x) + 2\cos (\frac{\pi}{2}x) -\pi x \sin (\frac{\pi}{2}x) \end{equation} which is \begin{equation} g'(x) = -\frac{\pi^2}{4} \cos (\frac{\pi}{2}x)(1-x^2) + 2\cos (\frac{\pi}{2}x) \end{equation} Factorize to get \begin{equation} g'(x) = [-\frac{\pi^2}{4}(1-x^2) + 2]\cos (\frac{\pi}{2}x) \end{equation} You could further factorize as \begin{equation} g'(x) = \big(x - \sqrt{1 - \frac{8}{\pi^2}} \big)\big(x + \sqrt{1 - \frac{8}{\pi^2}} \big)\cos (\frac{\pi}{2}x) \end{equation}

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On the interval $[0,1]$

In the interval $[0,1]$, we know that both $\cos(\frac{\pi}{2})$ and $x + \sqrt{1 - \frac{8}{\pi^2}}$ are positive. So in $[0,1]$ the sign of $g'(x)$ depends on $x - \sqrt{1 - \frac{8}{\pi^2}}$. Obviously we get that $g'(x) < 0$ in $[0,x_0]$ and $g'(x) > 0$ in $[x_0,1]$ where $x_0 = \sqrt{1 - \frac{8}{\pi^2}}$. So there is a minimum at $x_0$. But $g(0) = g(1) = 0$. So $g(x) < 0$ in $[0,1]$.

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On the interval $[1,2]$

In the interval $[1,2]$, we have that both $x + \sqrt{1 - \frac{8}{\pi^2}}$ and $x + \sqrt{1 + \frac{8}{\pi^2}}$ are positive so the sign of $g'(x)$ depends on $\cos \frac{\pi}{2}x$ which is negative in $[1,2]$. Hence, $g(x)$ is decreasing decreasing in $[1,2]$, but $g(1) = 0$, hence $g(x)$ is also negative in $[1,2]$.

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Concluding on $[0,2]$

We conclude that in $[0,2]$, \begin{equation} g(x) \leq 0 \end{equation} Hence $f(x)$ is decreasing in $[0,2]$, excluding $x = 1$.