The function $ f (x) = \frac{x^2 – 3x + 2}{x^2 + 2x – 3} $ is

Q: The function $\large f (x) = \frac{x^2 – 3x + 2}{x^2 + 2x – 3} $ is

A) max. at x = -3

(B) min. at x = -3 and max. at x = 1

(C) increasing in its domain

(D) none of these

Sol: $\large f (x) = \frac{x^2 – 3x + 2}{x^2 + 2x – 3} $

$\large f (x) = \frac{(x-1)(x-2)}{(x-1)(x+3)} , x \ne 1 , -3 $

$\large f (x) = \frac{(x-2)}{(x+3)}$

$\large \frac{df(x)}{dx} = \frac{(x+3)-(x-2)}{(x+3)^2}$

$\large = \frac{5}{(x+3)^2} > 0 \;  \forall x \ne 1 , -3$

Clearly f (x) is increasing in its domain.

Hence (C) is correct.