what-range-the-function-y-x-1-x-2-x-

Question Number 76919 by jagoll last updated on 01/Jan/20

w h a t r a n g e

t h e f u n c t i o n y = \frac{x - 1}{\sqrt{x^{2} + x}} ?

Answered by MJS last updated on 01/Jan/20

defined for - \infty < x < - 1 \lor 0 < x < + \infty

\frac{d}{d x} [\frac{x - 1}{\sqrt{x^{2} + x}}] = \frac{3 x + 1}{2 {(x^{2} + x)}^{3 / 2}} = 0 \Rightarrow x = - \frac{1}{3} \Rightarrow

\Rightarrow no real extremes

\lim_{x \to - \infty} \frac{x - 1}{\sqrt{x^{2} + x}} = - 1

\lim_{x \to - 1^{-}} \frac{x - 1}{\sqrt{x^{2} + x}} = - \infty

\lim_{x \to 0^{+}} \frac{x - 1}{\sqrt{x^{2} + x}} = - \infty

\lim_{x \to + \infty} \frac{x - 1}{\sqrt{x^{2} + x}} = 1

\Rightarrow range is - \infty < y < 1

Commented by jagoll last updated on 01/Jan/20

if y equal to 1 \Rightarrow \sqrt{x^{2} + x} = x - 1

squaring in sides both

x^{2} + x = x^{2} - 2 x + 1 \Rightarrow 3 x = 1

x = \frac{1}{3} satisfy in domain function

Commented by john santu last updated on 01/Jan/20

s o r r y s i r . s q u a r i n g i n s i d e s b o t h

w o r k i f x - 1 ⩾ 0 . t h e v a l u e x - 1 < 0 f o r x = \frac{1}{3}