Question 3.5.3: Let f and g be real functions, defined by f(x) = 1/(x + 4) a......
Let f and g be real functions, defined by f(x)=\frac{1}{(x + 4)} and g(x)=(x+4)^{3} .
Find (i)(f+g)(x)\qquad (ii) (f-g)(x) \qquad(iii) (f g)(x)\qquad (i v)\left(\frac{f}{g}\right)(x) \qquad (v) \left(\frac{1}{f}\right)(x)
Clearly, f(x)=\frac{1}{(x + 4)} is defined for all real values of x except that at which x+4=0 , i.e., x=-4 .
\therefore \quad \operatorname{dom}(f)=R-\{-4\} .
And, g(x)=(x+4)^{3} is defined for all x \in R . So, \operatorname{dom}(g)=R .
\therefore \quad \operatorname{dom}(f) \cap \operatorname{dom}(g)=R-\{-4\} \cap R=R-\{-4\} .
(i) (f+g): R-\{-4\} \rightarrow R is given by
(f+g)(x)=f(x)+g(x)=\frac{1}{(x + 4)}+(x+4)^{3}=\frac{1 + (x + 4)^{4}}{(x + 4)}.
(ii) (f-g): R-\{-4\} \rightarrow R is given by
(f-g)(x)=f(x)-g(x)=\frac{1}{(x + 4)}-(x+4)^{3}=\frac{1-(x + 4)^{4}}{(x + 4)}.
(iii) (f g): R-\{-4\} \rightarrow R is given by
(f g)(x)=f(x) \cdot g(x)=\frac{1}{(x + 4)} \times(x+4)^{3}=(x+4)^{2} .(iv) \{x: g(x)=0\}=\left\{x:(x+4)^{3}=0\right\}=\{x: x+4=0\}=\{-4\} .
\therefore \quad \operatorname{dom}\left(\frac{f}{g}\right)=\operatorname{dom}(f) \cap \operatorname{dom}(g)-\{x: g(x)=0\}=R-\{-4\} .\left(\frac{f}{g}\right): R-\{-4\} \rightarrow R is given by
\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{\frac{1}{(x + 4)}}{(x+4)^{3}}=\frac{1}{(x+4)^{4}}, x \neq-4 .(v) Clearly, f(x) \neq 0 for any x \in R-\{-4\} .
\therefore \quad \frac{1}{f}: R-\{-4\} \rightarrow R is given by
\left(\frac{1}{f}\right)(x)=\frac{1}{f(x)}=\frac{1}{\frac{1}{(x + 4)}}=(x+4).