Let $f, g$ : $R \to R$ be defined respectively by $f (x) = x + 1, g (x) = 2 x - 3$ . Find $f + g, f - g$ and $\frac{f}{g}$

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Solution

$f, g : R \to R$ is defined as
$f (x) = x + 1$
$g (x) = 2 x - 3$
Now, $(f + g) (x) = f (x) + g (x) = (x + 1) + (2 x - 3) = 3 x - 2$
$∴ (f + g) (x) = 3 x - 2$
Now, $(f - g) (x) = f (x) - g (x) = (x + 1) - (2 x - 3) = x + 1 - 2 x + 3 = - x + 4$
$∴ (f - g) (x) = - x + 4$
$(\frac{f}{g}) (x) = \frac{f (x)}{g (x)}, g (x) \neq 0, x \in R$
$∴ (\frac{f}{g}) (x) = \frac{x + 1}{2 x - 3}, 2 x - 3 \neq 0 o r 2 x \neq 3$
$∴ (\frac{f}{g}) (x) = \frac{x + 1}{2 x - 3}, x \neq \frac{3}{2}$

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