Let $f: R \rightarrow R: f(x)=2 x+5$ and $g: R \rightarrow R: g(x)=x^{2}+x$.
Find
(i) $(f+g)(x)$
(ii) $(f-g)(x)$
(iii) (fg) (x)
(iv) $(f / g)(x)$
(i) Given:
$f(x)=2 x+5$ and $g(x)=x^{2}+x$
(i) To find: $(f+g)(x)$
$(f+g)(x)=f(x)+g(x)$
$=(2 x+5)+\left(x^{2}+x\right)$
$=2 x+5+x^{2}+x$
$=x^{2}+3 x+5$
Therefore,
$(f+g)(x)=x^{2}+3 x+5$
(ii) To find: $(f-g)(x)$
$(f-g)(x)=f(x)-g(x)$
$=(2 x+5)-\left(x^{2}+x\right)$
$=2 x+5-x^{2}-x$
$=-x^{2}+x+5$
Therefore,
$(f+g)(x)=-x^{2}+x+5$
(iii) To find: (fg)(x)
$(f g)(x)=f(x) \cdot g(x)$
$=(2 x+5) \cdot\left(x^{2}+x\right)$
$=2 x\left(x^{2}\right)+2 x(x)+5\left(x^{2}\right)+5 x$
$=2 x^{3}+2 x^{2}+5 x^{2}+5 x$
$=2 x^{3}+7 x^{2}+5 x$
Therefore,
$(f g)(x)=2 x^{3}+7 x^{2}+5 x$
(iv) To find $:\left(\frac{f}{g}\right)(x)$
$\left(\frac{f}{g}\right)(x)=\left(\frac{f(x)}{g(x)}\right)$
$=\left(\frac{2 x+5}{x^{2}+x}\right)$
Therefore,
$\left(\frac{f}{g}\right)(x)=\left(\frac{2 x+5}{x^{2}+x}\right)$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.