Question

Let f(x) = 2x + 5 and $g\left(x\right)=x^2+x$. Describe
(i) f + g
(ii) f - g
(iii) $\frac{f}{g}$. Find the domain in each case.

Answer 1

We have,
$f\left(x\right)=2x+5$ and $g\left(x\right)=x^2+x$
We observe that $f\left(x\right)=2x+5$ is defined for all $xϵR$
So, domain (f) = R
Clearly $g\left(x\right)=x^2+x$ is defined for all $xϵR$
So, domain (g) = R
$\therefore$ Domain (f) $cap$ Domain (g) = R

(i) Clearly, $(f+g):R\to R$ is given by
(f + g) (x) = f(x) + g(x)
$=2x+5+x^2+x=x^2+3x+5$
Domain (f + g) = R

(ii) We find that $f-g:R\to R$ is defined as
$(f-g)\left(x\right)=f\left(x\right)-g\left(x\right)=2x+5-(x^2+x)=2x+5-x^2-x=-x^2+x+5$
Domain (f-g) = R

(iii) We find that $fg:R\to R$ is given by
$f\left(g\right)\left(x\right)=f\left(x\right)\times g\left(x\right)=(2x+5)\times (x^2+x)=2x^3+2x^2+5x^2+5x=2x^3+7x^2+5x$
Domain (fg) = R

(iv) We have,
$g\left(x\right)=x^2+x\therefore f\left(x\right)=0\Rightarrow x^2+x=0\Rightarrow x(x+1)=0\Rightarrow x=0\text{or},x=-1$
So, domain $\left(\frac{f}{g}\right)=\text{domain}\left(f\right)\cap \text{domain}\left(g\right)-\{x:g(x)=0\}=R-\{-\varphi ,0\}$
We find that, $\frac{f}{g}:R-\{-1,0\}\to R$ is given by $\left(\frac{f}{g}\right)\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}=\frac{2x+5}{x^2+x}\text{Domain}\left(\frac{f}{g}\right)=R-\{-1,0\}$