# Prove that

Question:

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) $(f / g)(x)$

Solution:

(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,

$\left(\frac{f}{g}\right)(x)=\left(\frac{2 x+5}{x^{2}+x}\right)$