Question 3.5.10: Find the product of the identity function by the modulus fun......

Let f: R \rightarrow R: f(x)=x and g: R \rightarrow R: g(x)=|x| be the identity function and the modulus function respectively.

Then, \operatorname{dom}(f g)=\operatorname{dom}(f) \cap \operatorname{dom}(g)=R \cap R=R .

Now, (f g)(x)=f(x) \cdot g(x)=x \cdot|x| \\\qquad \qquad \qquad =x \cdot\left\{\begin{array}{l}x, \text { when } x \geq 0 \\-x, \text { when } x<0\end{array}=\left\{\begin{array}{l}x^{2}, \text { when } x \geq 0 \\-x^{2}, \text { when } x<0 .\end{array}\right.\right.

Hence, (f g)(x)=\left\{\begin{array}{l}x^{2}, \text { when } x \geq 0 \\ -x^{2}, \text { when } x<0 .\end{array}\right.