If x is real, show that each of the following expression is capabale of assuming all real values: (i) (2x^2+4x+1)/(x^2+4x+2) (ii) p^2/(1-x)-q^2/(1+x)?

#(i) (2x^2+4x+1)/(x^2+4x+2) (ii) p^2/(1-x)-q^2/(1+x)#

1 Answer
Jan 18, 2018

I'm going to do this a little theoretically; I hope that's okay.

We can only see the sign of the function switch if we have an asymptote or a zero, so let's find those.

For the first, we find the vertical asymptotes, i.e. where the denominator is 0:

#x^2 + 4x + 2 = 0#
#x = (-4 pm sqrt(16 - 4(2)))/(2) = (-2 pm sqrt2) approx {-3.4, -0.6}#
The actual values aren't that important past the first decimal.

The zeroes are when the numerator is 0:

#2x^2+4x+1 = 0 #
#x = (-4 pm sqrt(16 - 4 * 2 * 1))/(2 * 2) = -1 pm sqrt2/2 approx {-1.71, -0.29}#

So we need to consider the signs in the following regions:

# (-infty, -3.4), (-3.4, -1.71), (-1.71, -0.6), (-0.6, -0.29), (-0.29, infty)#

If either end of the region is an asymptote, it will go to # pm infty# based on the sign of any point in that region.

Therefore, we can just check some numbers in these ranges:
#(-infty, -3.4): f(-5) = 31/7 > 0 #
(So the function goes to #+infty# as it approaches -3.4 negatively).

#(-3.4, -1.71): f(-2) = -1/2#
(So the function goes to #-infty# as it approaches -3.4 positively).

All of the above work was being really careful and most likely you don't have to be so careful.

For the second equation, we can simplify it into one statement
#(p^2(1+x) - q^2(1-x))/((1+x)(1-x)) =((p^2-q^2)+(p^2 + q^2)x )/((1+x)(1-x)) #

So, we know that the slope on the numerator is positive since it is the sum of two squares and the denominator has asymptotes at #pm 1#.

As the function approaches #x=1# positively, we have a positive numerator and a negative denominator, meaning it goes to negative infinity.

As the function approaches #x=-1# negatively, we have a negative numerator and a negative denominator, meaning a positive value, i.e. it goes to positive infinity.

Therefore, both functions can take on any real value.