Re: Finding set of possible values of a given function

On Feb 21, 9:45 am, newbar...@xxxxxxxxx wrote:
Hello,

I'm self teaching through a math's book and at the end of a chapter on
inequalities. They're currently being employed to find the possible
values of functions. I can do the exercises but I don't really
understand fully WHY it works!

The last question was asking about (x + 1) / (2x^2 + x + 1).

I converted that into a quadratic equation in 'x':

2y^2 + (y - 1)x + (y - 1) = 0

And the next step was to look for "real roots" of this quadratic;

i.e. that satisfy the inequality b^2 - 4ac >= 0.

It's at this stage where my understanding of the meaning behind the
process breaks down. I'm having trouble understanding the correlation
of real roots of the quadratic with the range of the function.

I thought the roots just showed you where the function intersected the
x-axis. I've seen graphs that don't intersect the x-axis at all (i.e.
graphs of f(x) = x^2 + 1). That doesn't have real roots, but it has an
output range!

I've been working with this for over a week now and even went over the
whole inequality chapter again but I'm still not "getting it"!. Hoping
for some great explanations.

Let F(x) = (x + 1) / (2x^2 + x + 1). Note first that the denominator D
(x) = 2x^2 + x + 1 is always > 0 for all x (because it is > 0 at x = 0
and it has no real roots). The denominator is > 0, while the numerator
N(x) = x + 1 changes sign at x = -1. Therefore, F(x) takes on both
positive and negative values. Note also that for large |x|, the
denominator is much, much larger than the numerator, so F(x) behaves
like 1/(2x). In fact, F(x) = x(1 + 1/x)/[2x^2 (1 + 1/(2x) + 1/
(2x^2) )] = (1/(2x)) *[factor that -> 1 as |x| goes to infinity]. In
other words, F(x) --> 0 as |x| --> infinity. Basically, this fact,
together with the fact that the numerator never vanishes, implies that
|F(x)| remains bounded, so F(x) itself remains between some lower
bound L < 0 and some upper bound U > 0. The achievable values (the
range of F) will be the interval L <= y <= U.

We can find L and U by your method, as follows. To find U, you want to
know the smallest value of y so that F(x) <= y always holds. At v = U,
the line y = v should just touch the graph y = F(x) at one point, so
we can look for a root of F(x) = v that is "single". Re-write the
equation F(x) = v as (x+1) = v*(2x^2 + x + 1), or (2v)x^2 + (v-1)x +
(v-1) = 0, whose roots are x = {(1-v) +- sqrt[(v-1)^2 - 4(2v)(v-1)]}/
(4v). Since we have a "single" root at the touching point, we need the
square root part to vanish, so we need (v-1)^2 -(8v)(v-1) = 0, giving
v = 1 or v-1 - 8v = 0, hence v = -1/7. The value v = 1 cannot be the
minimum because we know already that F(x) can have negative values;
therefore, the root v = 1 is the maximum (giving x = 0 in the root-
formula for x). The other root v = -1/7 actually gives the minimum,
because the argument for finding L is the same: the line y = v (v < 0)
must touch the graph y = F(x) at just one point when v = L. Thus, L =
-1/7, occurring at x = -2 (by plugging v = -1/7 in the root formula
for x).

The above method for finding L and U is somewhat ad hoc. A more
generally-applicable method is to use calculus, but I don't know if
you would be in a position to use such techniques. Anyway, if you know
some calculus, you can try to maximize and minimize F(x) by setting
the derivative to zero. The derivative of F is F'(x) = -2x*(x+2)/
(2x^2+x+1)^2, which vanishes at x = 0 and x = -2. We could perform a
second-derivative test to check which point is the maximum and which
the minimum, but we can reason it out more directly. F is < 0 for x <
-1 and is > 0 for x > -1. F has a maximum or minimum at x = 0, is > 0
there, while going to zero as x goes to + infinity. That means that x
= 0 cannot be the minimum (do you see why?); it must be the maximum.
Thus, U = F(0) = 1. It now follows that x = -2 must be the minimum
(actually, the minimizer) of F, so L = F(-2) = -1/7.

R.G. Vickson


Regards,

Pete

.

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