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Question:
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If $f$ and $g$ are two real valued functions defined as $f(x)$ $=2 x+1$ and $g(x)=x^2+1$, then find
(i) $f+g$
(ii) $f-g$
(iii) $f g$
(iv) $\frac{f}{g}$
(i) $f+g$
(ii) $f-g$
(iii) $f g$
(iv) $\frac{f}{g}$
Solution:
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Verified Answer
Given $f(x)=2 x+1$ and $g(x)=x^2+1$
(i) $(f+g)(x)=f(x)+g(x)$
$=2 x+1+x^2+1=x^2+2 x+2$
(ii) $(f-g)(x)=f(x)-g(x)=(2 x+1)-\left(x^2+1\right)$
$=2 x+1-x^2-1=2 x-x^2=x(2-x)$
(iii) $(f g)(x)=f(x) \cdot g(x)=(2 x+1)\left(x^2+1\right)$
$=2 x^3+2 x+x^2+1=2 x^3+x^2+2 x+1$
(iv) $\frac{f}{g}(x)=\frac{f(x)}{g(x)}=\frac{2 x+1}{x^2+1}$
(i) $(f+g)(x)=f(x)+g(x)$
$=2 x+1+x^2+1=x^2+2 x+2$
(ii) $(f-g)(x)=f(x)-g(x)=(2 x+1)-\left(x^2+1\right)$
$=2 x+1-x^2-1=2 x-x^2=x(2-x)$
(iii) $(f g)(x)=f(x) \cdot g(x)=(2 x+1)\left(x^2+1\right)$
$=2 x^3+2 x+x^2+1=2 x^3+x^2+2 x+1$
(iv) $\frac{f}{g}(x)=\frac{f(x)}{g(x)}=\frac{2 x+1}{x^2+1}$
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