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We know that : $$f(x)= sgn(x)$$ $$g(x) = \lfloor x \rfloor$$ $$h(x) = x$$ What is the graph of $f(g(x))$ and $f(g(x))\times h(x) $ ?

My try : I tried to draw the graph for several intervals but I'm really confused about it . I'll be happy if someone draw the graphs and explain it.

Thanks!

2 Answers 2

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Just compute the points on each interval.

$$\bbox[border:2px solid lightblue]{f(g(x))} \quad \quad \quad\quad \quad\quad \quad\quad \quad\quad \quad \bbox[border:2px solid orange]{f(g(x))h(x)} $$

enter image description here

Notice that for $f(g(x))$, It's easily seen (from the definition of the functions) that they won't get out of the interval $[1,-1]$ but as you multiply it by $h$, they will. Also, the definition of $sgn(x)$ I am using is:

$$sgn (x)=\begin{cases} -1 & \text{if } x < 0, \\ 0 & \text{if } x = 0, \\ 1 & \text{if } x > 0. \end{cases}$$

I am warning because I have already seen some variations of it. First you compute the function for the innermost values: That is: $g(x_0)$, then you compute $f(g(x_0))$ and then you multiply by $h(x)$. I you draw the blue function first, you can use the previous graph to make the next one.

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    @S.H.W Just added. – 2017-01-07
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    Are you sure about $f(g(x))h(x)$ ? – 2017-01-07
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    I think it is wrong – 2017-01-07
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    @S.H.W Why you think that? – 2017-01-07
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    @Rohan $f(x)<0$ for $x<0$. $h(x)<0$ for $x<0$. $(-a)(-b)=?$ – 2017-01-07
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    @OppaHilbertStyle Sorry. It was my mistake. – 2017-01-07
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    @Rohan No problem. – 2017-01-07
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Hint: $$\operatorname{sgn}(\lfloor x \rfloor ) =\begin {cases} -1 & \text { if } x < 0,\\ 0 & \text { if } x\in [0,1),\\ 1 & \text { if } x \geq 1. \end {cases} $$

Use this hint to find the values of $x\operatorname{sgn} (\lfloor x\rfloor ) $ at various values of $x $. Hope it helps.