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

2 Answers
May 22, 2015

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?

May 23, 2015

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)

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