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?
