Find the points of discontinuity of the graph and also draw the graph in rough

Question

$f(x)=\begin{cases} sin(\dfrac{\pi}{4}(x-[x])) & \text{for }[x] \space is \space odd, x\ge0\\ cos(\dfrac{\pi}{4}(1-x+[x])) & \text{for }[x] \space is \space even,x\ge0 \end{cases}$

where $[x]$ is the largest integer smaller than or equal to x

(i)Sketch the graph of $f(x)$ on plain paper

(ii)Determine the points of dicontinuities of $f$ and the points where $f$ is not differentiable

My attempt

The graph looks like a wave where there is no negative half and it reaches its crest at every even integer and reaches 0 at every odd integer starting from 0 as $x\ge0$ i.e it alternates between $0 \space \text{&} \dfrac{1}{\sqrt{2}}$

At every integer where $[x]$ is either even or odd, it is discontinuous and not differentiable

Am I right? I cannot take a snapshot of the graph as many users do not advocate in favor of it. If you insist I can do that for you. Please draw the graph and let me know whether I am wrong or right.

Answer 1

Hint

As you noticed the only points where there is something to investigate are $x\in \Bbb Z$.

• For $x=(2n+1)^+$ you have $[x]=2n+1$ and: $$f(x)=\sin \left(\frac{\pi}{4}((2n+1)^+-(2n+1)) \right)=\sin(0)=0$$ For $x=(2n+1)^-$ you have $[x]=2n$ and: $$f(x)=\cos \left(\frac{\pi}{4}\left(1-(2n+1)^-+2n\right) \right)=\cos(0)=1$$
• For $x=(2n)^+$ you have $[x]=2n$ and: $$f(x)=\cos \left(\frac{\pi}{4}\left(1-(2n)^++2n\right) \right)=\cos\left(\frac{\pi}{4} \right)=\frac{1}{\sqrt{2}}$$ For $x=(2n)^-$ you have $[x]=2n-1$ and: $$f(x)=\sin \left(\frac{\pi}{4}\left((2n)^--(2n-1)\right) \right)=\sin\left(\frac{\pi}{4} \right)=\frac{1}{\sqrt{2}}$$

So in fact $f$ is continuous on even integers.

To study if $f$ is differentiable you also have to study $f$ near the integers the same way.

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Here is the plot of the function I have obtained: