# Question about domain of a function

My question is from Apostol's Vol. 1 One-variable calculus with introduction to linear algebra textbook.

Page 56. Exercise 6. Let $$f$$ be defined as follows: $$f(x)=1$$ for $$0\le x\le1$$; $$f(x)=2$$ for $$1\lt x\le2$$. The function is not defined if $$x\lt0$$ or if $$x>2$$.

$$a)$$ Draw the graph of $$f$$.

$$b)$$ Let $$g(x)=f(2x).$$ Describe the domain of $$g$$ and draw its graph.

$$c)$$ Let $$h(x)=f(x-2).$$ Describe the domain of $$h$$ and draw its graph.

$$d)$$ Let $$k(x)=f(2x)+f(x-2).$$ Describe the domain of k and draw its graph.

The attempt at a solution: I drew graph of $$f$$, it would look like this (Except function is not defined when $$x < 0$$ or $$x > 2$$): As for $$b)$$ and $$c)$$, since $$g(x)=f(2x)$$, on interval $$0\le x\le1$$, $$g(x)=2$$ as I understand, and on interval $$1\lt x\le2$$, $$g(x)=4$$, but since function $$f$$ is defined from $$x=0$$ to $$x=2$$, wouldn't same be true for $$g(x)$$? Wouldn't same be true for $$c)$$? What am I missing? because answer says that domain of function of $$g(x)$$ and $$h(x)$$ is different from domain of $$f$$, and domain of $$k(x)=f(2x)+f(x-2)$$ is empty.

Notice that for $(b)$

$$0\leq 2x \leq 1 \Rightarrow 0 \leq x \leq \frac{1}{2}$$ and $$1 \leq 2x \leq 2 \Rightarrow \frac{1}{2}\leq x \leq 1$$

The domain of $g$ is $\Big[0,\frac{1}{2}\Big] \cup \Big[\frac{1}{2},1\Big] = [0,1]$.

Similar goes for $(c)$. As for $(d)$ we have

$$\Big[0,\frac{1}{2}\Big]\cap\Big[\frac{1}{2},1\Big]\cap[2,4]\cap[3,5] = \emptyset$$

• Thank you very much for help, as you can see, sadly my algebra knowledge sucks. – George Apriashvili Nov 2 '14 at 16:15
• @GeorgeDirac I'm glad I could help. And it's normal at first, later on you will get used to the ideas and will be able to work on your own. – Aaron Maroja Nov 2 '14 at 16:17

For the function $g(x) = f(2x)$, let $t = 2x$. Then $x = t/2$. We know that

$$f(t) = \begin{cases} 1 & \text{if 0 \leq t \leq 1}\\ 2 & \text{if 1 < t \leq 2} \end{cases}$$ Replacing $t$ by $2x$ yields $$f(2x) = \begin{cases} 1 & \text{if 0 \leq 2x \leq 1}\\ 2 & \text{if 1 < 2x \leq 2} \end{cases}$$ Thus, $$g(x) = \begin{cases} 1 & \text{if 0 \leq x \leq \dfrac{1}{2}}\\ 2 & \text{if \dfrac{1}{2} < x \leq 1} \end{cases}$$ so the domain of $g$ is $[0, 1]$, which is obtained from the domain of $f$ by dividing each element in the domain of $f$ by $2$.

For the function $h(x) = f(x - 2)$, let $t = x - 2$. Then $x = t + 2$. Thus, the domain of $h$ is $[2, 4]$, which is obtained by adding $2$ to each element in the domain of $f$.

The domain of $k(x)$ is the intersection of the domains of $g(x)$ and $h(x)$.

Note that the transformation $x \to 2x$ stretches the graph horizontally by a factor of $1/2$, while the transformation $x \to x - 2$ shifts the graph horizontally to the right by $2$ units.

• Thanks for detailed answer. I understand now. – George Apriashvili Nov 2 '14 at 16:22