Finding Range Method

[anja126]

Hi everyone

It's about the equation below. I want to find the range of the function, but I don't know how. The shown solution is way to complicated for me, so if anyone could explain or show a easiert method, that would be much appreciated.

I know that x^2-x+1 can't be zero for all x and that therefore the domain is not limited

Thank you

 

[Dr.Peterson]

It's about the equation below. I want to find the range of the function, but I don't know how. The shown solution is way to complicated for me, so if anyone could explain or show a easier method, that would be much appreciated.

I know that x^2-x+1 can't be zero for all x and that therefore the domain is not limited

Thank you

View attachment 22206

What part of it is too hard for you?

I think this is probably the easiest method; they are solving a rational equation, using the discriminant to determine the values of t for which it has a solution. (It could probably have been explained a little more clearly (that is, using less notation and more words.)

So we'll have to work with you so that you can carry this out. Please show us where you fail to understand, and we can discuss that part.

 

[pka]

Have a look at this LINK

 

[JeffM]

I agree that the answer provided is correct,

But [MATH]0 \le t \le \dfrac{4}{3}[/MATH]
does not entail that t ever equals 4/3. I do not yet see why the method used will necessarily give the right answer in all cases.

I repeat that I do not disagree in this case because I can confirm the answer using calculus.

[MATH]f(x) = \dfrac{x^2}{x^2 - x + 1}.[/MATH]
I agree that the domain of f(x) is all of R.

[MATH]f'(x) = \dfrac{2x(x^2 - x + 1) - x^2(2x - 1)}{(x^2 - x + 1)^2} = \dfrac{-x^2 + 2x}{(x^2 - x +1)^2}.[/MATH]
[MATH]f'(x) = 0 \iff -x^2 + 2x = 0 \iff x = 0 \text { or } x = 2.[/MATH]
That gives the location of the extrema. Obviously f(0) = 0. So that is confirmed.

[MATH]f(2) = \dfrac{4}{4 - 2 + 1} = \dfrac{4}{3}.[/MATH]
So 4/3 is confirmed.

Is my uncertainty justified?

 

[Dr.Peterson]

I agree that the answer provided is correct,

But [MATH]0 \le t \le \dfrac{4}{3}[/MATH]
does not entail that t ever equals 4/3. I do not yet see why the method used will necessarily give the right answer in all cases.

Is my uncertainty justified?

I see no problem in their work, though of course it's reasonable to add a check at the end.

They've shown that the discriminant will be non-negative for all values of t in the indicated interval, which should be sufficient.

 

[JeffM]

I see no problem in their work, though of course it's reasonable to add a check at the end.

They've shown that the discriminant will be non-negative for all values of t in the indicated interval, which should be sufficient.

I see no problem in their working. My sole question is how they show that the extreme values are actually achieved. My point is that t is a dependent variable.

I suppose you could solve the extreme values of t for x to determine that an x exists that generates those extreme values, but that they do not do.

Anyway, thanks for considering the question.

 

[Dr.Peterson]

My sole question is how they show that the extreme values are actually achieved. My point is that t is a dependent variable.

I suppose you could solve the extreme values of t for x to determine that an x exists that generates those extreme values, but that they do not do.

What they are doing amounts to finding the inverse relation, and determining its domain. (At that point, t is in fact the independent variable!) For all t in the interval found, there is a corresponding real value of x, for which a formula is given. This includes the endpoints, at which the discriminant is zero, so that there is one value of x, namely t/[2(t-1)] = 0 or 2 for t = 0 or 4/3 respectively. They rightly show double arrows (iff) at every step. They don't need to actually calculate those values, as the work shows they exist.

 

[JeffM]

What they are doing amounts to finding the inverse relation, and determining its domain. (At that point, t is in fact the independent variable!) For all t in the interval found, there is a corresponding real value of x, for which a formula is given. This includes the endpoints, at which the discriminant is zero, so that there is one value of x, namely t/[2(t-1)] = 0 or 2 for t = 0 or 4/3 respectively. They rightly show double arrows (iff) at every step. They don't need to actually calculate those values, as the work shows they exist.

Got it. Thanks. Now all we need to do is explain it to the OP.

Calculus approach is more straight forward in my opinion, but both are sound.

 

[Dr.Peterson]

Now all we need to do is explain it to the OP.

Calculus approach is more straight forward in my opinion, but both are sound.

Yes. I think they did a wonderful job of doing the work, but a horrible job of communicating it to anyone who is not already a good mathematician, comfortable with all the notation, and not needing any explanation! As I said initially, "It could probably have been explained a little more clearly (that is, using less notation and more words)." That was an intentional understatement.

And, of course, it seems likely that this is not a calculus course.

@anja126, are you following this? Can you give us any more input?

 

[anja126]

@JeffM your method actually makes more sense to me than the one they provided, thank you

 

[anja126]

What part of it is too hard for you?

I think this is probably the easiest method; they are solving a rational equation, using the discriminant to determine the values of t for which it has a solution. (It could probably have been explained a little more clearly (that is, using less notation and more words.)

So we'll have to work with you so that you can carry this out. Please show us where you fail to understand, and we can discuss that part.

I don't understand why they use the discriminant to calculate the range of f(x). And I don't get the end part, where they calculate the values 0 and 4/3 and put it between the greater or equal as arrows. Would I just need to put t in between and make it greater or equal as 0 and smaller or equal as 4/3, that's it?

 

[Dr.Peterson]

I don't understand why they use the discriminant to calculate the range of f(x). And I don't get the end part, where they calculate the values 0 and 4/3 and put it between the greater or equal as arrows. Would I just need to put t in between and make it greater or equal as 0 and smaller or equal as 4/3, that's it?

The range of [MATH]f[/MATH] is the set of possible values of [MATH]t[/MATH]. Therefore, we want to find for what values of [MATH]t[/MATH] the value of [MATH]x[/MATH] that they obtained is a real number.

This is found by setting the discriminant (that is, the radicand in the expression they obtained) greater than or equal to 0. If this is less than zero, then there are no real solutions for [MATH]x[/MATH].

So now we have to solve the inequality [MATH]t^2 - 4(t-1)t \ge 0[/MATH]. The standard way to solve a polynomial inequality is to factor it and find which intervals between zeros satisfy the inequality. Expanding the LHS, we have [MATH]-3t^2+ 4t\ge 0[/MATH], which factors as [MATH]t(4-3t)\ge 0[/MATH]. This divides the number line into three regions at 0 and 4/3. For [MATH]t<0[/MATH], the first factor is negative and the second is positive, so the LHS is less than zero. For [MATH]t>4/3[/MATH], the first factor is positive and the second is negative, so the LHS is again less than zero. But for [MATH]0\le t\le 4/3[/MATH], both factors are positive and the inequality is satisfied. This is the range.

 

[JeffM]

As you can see from my replies, I was initially not even sure this made sense. And of course my method is only available if you know differential calculus. But it is a valid algebraic method.

The first thing that is a little confusing is the use of t rather than y.

[MATH]y = f(x) = \dfrac{x^2}{x^2 - x + 1}.[/MATH]
It is easy to show that f(x) is a real number for all real x because the denominator is everywhere positive.

Moreover, it is easy to show that f(x) is positive for any real, non-zero x and f(0) = 0. So they do not really need to worry about the bottom of the range, but they want to show that the method works for both the bottom and the top of the range.

Let's pick an arbitrary value for x. I'll call it a for arbitray. Because the domain is every real number, we know that f(a) is a real number. Call it r. (I'm ignoring their labels.)

[MATH]r = f(a) = \dfrac{a^2}{a^2 - a + 1}.[/MATH]
We see that we can turn this into a quadratic.

[MATH]a^2r - ar + r = a^2 \implies (r - 1)a^2 - ra + r.[/MATH]
We know that for a non-trivial quadratic to have any real root, its discriminant in the quadratic formula must be non-negative, right? So we ask ourselves whether that imposes any limits on r.

But before we can use the quadratic formula, we must make sure that the coefficient of a^2 is non-zero. But if r = 1, then
(r - 1) = 0. Is there any value of a that results in r = 1?

Now they take a clever shortcut to answer that question, but the obvious way to go is

[MATH]1 = \dfrac{x^2}{x^2 - x + 1} \implies x^2 - x + 1 = x^2 \implies x = 1.[/MATH]
So we have determined that 1 is in the range. And we have dealt with the case of r = 1. So we ask whether r can take on other values. And now we can use the quadratic formula.

[MATH]a = \dfrac{r \pm \sqrt{r^2 - 4(r - 1)r}}{2(r - 1)}.[/MATH]
If that discriminant is negative, a is not real. But we are only interested in real values for a.

[MATH]r^2 - 4(r - 1)r = r^2 - 4r^2 + 4r = 4r - 3r^2 = r(4 - 3r) \ge 0.[/MATH]
If r is negative, then r(4 - 3r) is negative. So r cannot be negative. If r is zero, that is ok. What value of a does that give us. Why a = 0, with f(0) = 0, which we know is the bottom of the range. If r > 0, then it must be true that

[MATH]4 - 3r \ge 0 \implies \dfrac{4}{3} \ge r.[/MATH]