What is the range of #y=[(1-x)^(1/2)]/(2x^2+3x+1)#?

First let us consider the domain:

For what values of #x# is the function defined?

The numerator #(1-x)^(1/2)# is only defined when #(1-x) >= 0#. Adding #x# to both sides of this you find #x <= 1#.

We also require the denominator to be non-zero.
#2x^2+3x+1 = (2x+1)(x+1)# is zero when #x = -1/2# and when #x = -1#.

So the domain of the function is

#{x in RR: x <= 1 and x != -1 and x != -1/2}#

Define #f(x) = (1-x)^(1/2)/(2x^2+3x+1)# on this domain.

Let us consider each continuous interval in the domain separately:

In each case, let #epsilon > 0# be a small positive number.

Case (a) : #x < -1#

For large negative values of #x#, #f(x)# is small and positive.
At the other end of this interval, if #x = -1 - epsilon# then

#f(x) = f(-1-epsilon) ~= sqrt(2)/(((2 xx -1)+1)(-1 - epsilon + 1))#

#= sqrt(2)/epsilon -> +oo# as #epsilon -> 0#

So for #x < -1# the range of #f(x)# is #(0, +oo)#

Case (b) : #-1/2 < x <= 1#

#f(-1/2+epsilon) ~= sqrt(3/2)//((2(-1/2+epsilon) + 1)(-1/2+1)#

#= sqrt(3/2)/epsilon -> +oo# as #epsilon -> 0#

#f(1) = 0/1 = 0#

So for #-1/2 < x <= 1# the range of #f(x)# is #[0, +oo)#

Case (c) : #-1 < x < -1/2#

#f(-1+epsilon) ~= sqrt(2)/(((2xx-1) + 1)(-1+epsilon+1))#

#= -sqrt(2)/epsilon -> -oo# as #epsilon -> 0#

#f(-1/2-epsilon) ~= sqrt(3/2)/((2(-1/2-epsilon) + 1)(-1/2+1)#

#= -sqrt(3/2)/epsilon -> -oo# as #epsilon -> 0#

So the interesting question is what is the maximum value of #f(x)# in this interval. To find the value of #x# for which this occurs look for the derivative to be zero.

#d/(dx)f(x)#
#= (1/2(1-x)^(-1/2)xx-1)/(2x^2+3x+1) + ((1-x)^(1/2)xx-1xx(2x^2+3x+1)^(-2)xx(4x+3))#

#= (-1/2(1-x)^(-1/2))/(2x^2+3x+1)-((1-x)^(1/2)(4x+3))/(2x^2+3x+1)^2#

#= ((-1/2(1-x)^(-1/2)(2x^2+3x+1))-((1-x)^(1/2)(4x+3)))/(2x^2+3x+1)^2#

This will be zero when the numerator is zero, so we would like to solve:

#-1/2(1-x)^(-1/2)(2x^2+3x+1)-((1-x)^(1/2)(4x+3)) = 0#

Multiply through by #2(1-x)^(1/2)# to get:

#-(2x^2+3x+1)-2(1-x)(4x+3) = 0#

That is:

#6x^2-5x-7 = 0#

which has roots #(5+-sqrt(25+4xx6xx7))/12 = (5+-sqrt(194))/12#

Of these roots, #x = (5-sqrt(194))/12# falls in the interval concerned.

Substitute this back into #f(x)# to find the maximum of #f(x) in this interval (approximately -10).

This seems over complex to me. Have I made any errors?

Answer: The range of the function is #(-oo, -10.58] uu [0,oo)#

For #x in (-oo, -1)# #-># #y in (0, oo)#
For #x in (-1, -0.5)# #-># #y in (-oo, -10.58]#
For #x in (-0.5, 1]# #-># #y in [0, oo)#