Increasing and decreasing function in certain interval

Click For Summary

Homework Help Overview

The problem involves analyzing the behavior of functions P(x) and Q(x), where P(x) is increasing and Q(x) is decreasing over a specified interval. The task is to determine the bounds m and M for a function γ(x) defined in various forms based on P(x) and Q(x). The context includes considerations of the first quadrant and the implications of differentiating the functions.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the implications of differentiating γ(x) for various cases, questioning the correctness of proposed bounds for m and M. There is uncertainty regarding the behavior of γ(x) in case (b), with suggestions to explore logarithmic transformations to clarify the relationship between P(x) and Q(x).

Discussion Status

Some participants express agreement on the correctness of the bounds for cases (a), (c), (d), and (e), while others question the ability to solve case (b) without additional information. The discussion reflects a lack of consensus on the implications of the first quadrant assumption and its impact on the analysis.

Contextual Notes

There is a mention that the original question did not specify the functions being in the first quadrant, leading to a reevaluation of the ability to answer certain parts of the problem.

harimakenji
Messages
89
Reaction score
0

Homework Statement


P(x) is an increasing function and Q(x) is a decreasing function in interval a ≤x≤b, x is positive. P(x) and Q(x) are located in 1st quadrant. Another function γ(x) satisfies m≤γ(x)≤M.
Find the value of m and M if:
a. γ(x) = P(x) - Q(x)
b. γ(x) = P(x) . Q(x)
c. γ(x) = [P(x)]2 – [Q(x)]2
d. γ(x) = 1/P(X) +Q(x)
e. γ(x)= (P(x))/(Q(x))-(Q(x))/(P(x))



Homework Equations


Maybe differentiation


The Attempt at a Solution


P(x) is increasing function = P'(x) is positive and P(a) < P(b)
Q(x) is decreasing function = Q'(x) is negative and Q(a) > Q(b)

a. γ'(x) = P'(x) - Q'(x). Since Q'(x) is negative, γ'(x) will be positive so γ(x) is increasing function.
m = P(a) - Q(b) and M = P(b) - Q(b). Is this correct?

b. γ'(x) = P'(x).Q(x) + P(x).Q'(x). I can not determine whether γ'(x) is positive or negative so I don't understand how to find m and M

c. γ'(x) = 2 P(X) P'(x) - 2 Q(x) Q'(x). The value of γ'(x) is positive so m = P2(a) - Q2(a) and M = P2(b) - Q2(b). Is this correct?

d. γ'(x) = -P'(x) / P2(x) + Q'(x). The value of γ'(x) is negative so m = 1/P(b) + Q(b) and M = 1/P(a) + Q(a). Is this correct?

e. γ'(x) = \frac{P&#039;(x).Q(x)-P(x).Q&#039;(x)}{Q^{2}(x)} - \frac{Q&#039;(x).P(x)-P&#039;(x).Q(x)}{P^{2}}. The value of γ'(x) is positive so m = (P(a))/(Q(a))-(Q(a))/(P(a)) and M = (P(b))/(Q(b))-(Q(b))/(P(b)). Is this correct?

Please help me to verify my answer and guide me how to solve (b). Thank you very much.
 
Physics news on Phys.org
hi harimakenji! :smile:

your a c d and e look fine

i don't see how you can solve b … it could be anything (wcihi you can see more clearly by taking logs: A = log P, B = logQ, so A is any increasing function, B is any decreasing function, and you want max and min of A + B)
 
tiny-tim said:
hi harimakenji! :smile:

your a c d and e look fine

i don't see how you can solve b … it could be anything (wcihi you can see more clearly by taking logs: A = log P, B = logQ, so A is any increasing function, B is any decreasing function, and you want max and min of A + B)

Hi tiny-tim. Thank you for the response

I don't really get what you are trying to tell me by using the logs. And one more thing, actually the real question doesn't mention anything about P(x) and Q(x) being in the first quadrant. I just added it myself.

If there is no information about the first quadrant, I think we can only answer questions (a) and (d). Do you have the same opinion?

Thank you very much
 
hi harimakenji! :smile:
harimakenji said:
If there is no information about the first quadrant, I think we can only answer questions (a) and (d). Do you have the same opinion?

yes

(and forget the logs, it doesn't matter)
 
tiny-tim said:
hi harimakenji! :smile:


yes

(and forget the logs, it doesn't matter)

Ok, I will forget about the logs for now.

Thank you very much for your help
 

Similar threads

Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K