Question

find the range of f(x)=3/2-x sq

Answer 1

f(x ) = 3/( 2 - x²)

y = 3/( 2 - x²)

2y -yx² = 3

x²y -2y + 3 = 0

x² =(2y -3)/y

x = ±√{( 2y -3)/y }

f( y) = ±√{(2y -3)/y}

domain of f( y) = range of f(x )

(2y -3)/y ≥0 ,

y ≥ 3/2 and y <0 and y ≠ 0

if y = 0

f(x ) = y = 3/(2 -x²) =0
3 ≠ 0

so, y ≠ 0

hence , range of f(x) € [ 3/2, ∞) U(-∞, 0)