Real Solutions for Real Values: (A,B,C,D)

  • Thread starter erisedk
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In summary: Since a, b, c are in arithmetic progression, we can change the terms in the above equation to get:##y^2 + 2y(b-c) + ... ≥ 0##This has a positive discriminate, so both roots are real. The roots of this quadratic are the two values of x for which the function crosses the x-axis. In the first case, the function is positive for x < a and negative for b < x < c. In the second case, the function is negative for x < a and positive for b < x < c. In the third case, the function is positive for x < b and negative for c < x.In summary, the function will have two
  • #1
erisedk
374
7

Homework Statement


For real ##x##, the function ##\dfrac{(x-a)(x-b)}{x-c}## will assume all real values provided

(A) ##a > b > c##
(B) ##a < b < c##
(C) ##a > c > b##
(D) ##a < c < b##

Homework Equations

The Attempt at a Solution


I'm really stuck on this. I know that ##x = c## is a vertical asymptote. I also know that I could substitute in values for ##a, b## and ##c## and graph it, but I don't want to do that. Again, I'm really not sure how to go about it. Please help.
 
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  • #2
erisedk said:

Homework Statement


For real ##x##, the function ##\dfrac{(x-a)(x-b)}{x-c}## will assume all real values provided

(A) ##a > b > c##
(B) ##a < b < c##
(C) ##a > c > b##
(D) ##a < c < b##

Homework Equations

The Attempt at a Solution


I'm really stuck on this. I know that ##x = c## is a vertical asymptote. I also know that I could substitute in values for ##a, b## and ##c## and graph it, but I don't want to do that. Again, I'm really not sure how to go about it. Please help.
Where are the x-intercepts?
 
  • #3
a and b.
 
  • #4
erisedk said:

Homework Statement


For real ##x##, the function ##\dfrac{(x-a)(x-b)}{x-c}## will assume all real values provided

(A) ##a > b > c##
(B) ##a < b < c##
(C) ##a > c > b##
(D) ##a < c < b##

Homework Equations

The Attempt at a Solution


I'm really stuck on this. I know that ##x = c## is a vertical asymptote. I also know that I could substitute in values for ##a, b## and ##c## and graph it, but I don't want to do that. Again, I'm really not sure how to go about it. Please help.

Isn't the question will assume positive real values?
 
  • #5
PeroK said:
Isn't the question will assume positive real values?
No. It makes sense as written.

It's saying that the range should include all real numbers, (-∞, ∞) .@erisedk ,
What is the is the behavior of the function as x → ± ∞ ?
 
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  • #6
As x → +∞ f(x) → ∞
and as x → -∞ f(x) → -∞
 
  • #7
erisedk said:
As x → +∞ f(x) → ∞
and as x → -∞ f(x) → -∞
Good.

Now starting with a negative at the left, in what order do you want two intercepts and a vertical asymptote ?
 
  • #8
1. There are multiple answers to this problem.

2. I understand what you're saying, we'll get these sort of hyperbola type graphs on both sides of the vertical asymptote, and on opposite sides of an asymptotes, the infinities that a function approaches are opposite (i.e + and - ∞)

3. I expanded the function and differentiated it to find the points of maxima/minima. I got a quadratic for the roots of f'(x) that has the discriminant ##4(c^2 - ac - bc + ab)##. Now, I'm not sure if I'm supposed to figure out what sign this is for the number of roots, because it seems complicated to work out for different sets of ##a, b## and ##c##.

4. Other than that, assuming that the discriminant is positive(for no reason, just assuming) so two roots, we'll have to figure out the maximum value of the function on the left side of ##c## and minimum value to the right of ##c## and compare them. Finding the roots of the quadratic for the maxima and minima seems rather arduous.

I don't really know what to do now.
 
  • #9
The substitution ##y = x-c## might be a good start, with the aim of transforming to as simple a function as possible.
 
  • #10
This is taking too much time, I suppose I should abandon the question at this point, especially since I've at least understood the key ideas.
 
  • #11
Thank you everyone for your help!
 
  • #12
Use a sign graph.

There are three critical points: x = a, x = b, and x = c. The rational function in question will change sign at each of these (none are equal in the choices) as x increases.

Trace the function from left to right (increasing x.) There are three possibilities. What does the graph look like for each?

  1. x - intercept, x - intercept, vertical asymptote
  2. x - intercept, vertical asymptote, x - intercept
  3. vertical asymptote, intercept, x - intercept
 
  • #13
SammyS said:
Use a sign graph.

There are three critical points: x = a, x = b, and x = c. The rational function in question will change sign at each of these (none are equal in the choices) as x increases.

Trace the function from left to right (increasing x.) There are three possibilities. What does the graph look like for each?

  1. x - intercept, x - intercept, vertical asymptote
  2. x - intercept, vertical asymptote, x - intercept
  3. vertical asymptote, intercept, x - intercept
Yeah, I did that. It doesn't solve the complete problem because there still is the question of whether the maxima on the left and minima on the right of the asymptote have some space between them. Furthermore, in case 2, we wouldn't get maximas or minimas, we'd get inflection points(or perhaps not even inflection points, not exactly sure there, that would happen if the discriminant of the derivative is negative).
 
  • #14
erisedk said:
Yeah, I did that. It doesn't solve the complete problem because there still is the question of whether the maxima on the left and minima on the right of the asymptote have some space between them. Furthermore, in case 2, we wouldn't get maximas or minimas, we'd get inflection points(or perhaps not even inflection points, not exactly sure there, that would happen if the discriminant of the derivative is negative).
In the first and third cases, you can't be sure regarding there being a gap in the range, without more analysis, which I know you're trying to avoid.

However you can be sure that in case 2, the entire range is covered.

There are not inflection points. That may be difficult to see. The second derivative is messy, but we are in the pre-calculus forum so ...
 
  • #15
I think I figured it out!
##y = \dfrac{(x-a)(x-b)}{x-c} ##
##(x - c) y = x^2 - (a+b)x + ab##
## x^2 - (a+b+y)x + ab + cy = 0##
Since x is real
D ≥ 0
## (a+b+y)^2 - 4(ab + cy) ≥ 0##
##y^2 + 2y(a+b-2c) + (a-b)^2 ≥ 0##
So, now discriminant of the quadratic in y should be negative as this quadratic is always positive
##4(a+b-2c)^2 - 4(a-b)^2 ≤ 0##
## (a + b - 2c + a - b)(a+b -2c - a +b)≤0##
##4(a-c)(b-c)≤0##
Ignoring the equality in ≤
## a - c<0 ## and ## b - c > 0## OR ## a - c > 0 ## and ## b -c < 0##
## a < c < b ## OR ## a > c > b##
 
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  • #16
erisedk said:
I think I figured it out!
...
That's quite interesting !

A few picky issues remain such as the = of the ≥ , etc.
 
  • #17
Yeah, I just chose to ignore it.
 
  • #18
You could analyze this function in terms of its slant asymptote.

Using long division or some other method, such as Mark44's decomposition, we get that
## \dfrac{(x-a)(x-b)}{x-c} = x + c-(a+b)+(c-a)(c-b)\dfrac{1}{x-c}##

## \quad = x + G+H\dfrac1{x-c}##​
for constants, G and H. Which have values:
##H=(c-a)(c-b)\ ##: Very important
##G=c-(a+b)\ ##: Not so important for this​

We can view our function as the sum of two basic functions; a linear function with slope of 1, and a reciprocal function, shifted horizontally, and having a coefficient, H, which could be either positive or negative. It's the sign of H that is crucial for the problem in this thread.
 
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