# Find the range

• June 23rd 2009, 05:20 AM
cloud5
Find the range
1. $y=\frac{3x}{x^2+1}$

2. $y=\frac{3x-5}{x^2-1}$

1. $-\frac{3}{2} \leq y \leq \frac{3}{2}$

2. $y \leq \frac{1}{2},y \geq \frac{9}{2}$
• June 23rd 2009, 05:34 AM
Quote:

Originally Posted by cloud5
1. $y=\frac{3x}{x^2+1}$

2. $y=\frac{3x-5}{x^2-1}$

1. $-\frac{3}{2} \leq y \leq \frac{3}{2}$

2. $y \leq \frac{1}{2},y \geq \frac{9}{2}$

(1) yx^2+y=3x

yx^2-3x+y=0

b^2-4ac>=0 .. provided that x is real .

(-3)^2-4y^2>=0

y^2<=9/4

then u will get -\sqrt(9/4)<=y<=sqrt(9/4)

same goes to the other one .