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?
