# A weird function i guess

1. Mar 6, 2009

### songoku

a weird function i guess....

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. The attempt at a solution

i don't know how to start......

2. Mar 6, 2009

### yyat

Re: a weird function i guess....

Since P(x) is an increasing function, what inequalities can you form with P(x),P(a) and P(b)? And with Q(x),Q(a) and Q(b)?

3. Mar 6, 2009

### songoku

Re: a weird function i guess....

i think i can get P(a) ≤ P(x) ≤ P(b) and Q(b) ≤ Q(x) ≤ Q(a)

but i'm confused because i don't know the value of P(a), P(b), Q(b), Q(a). it can be positive or negative.

if P(x) is exponential function Q(x) is linear, then the maximum value can be P(b).Q(b), if Q(b) is positive
but if the value of Q(b) is negative, then P(b).Q(b) can be the minimum value

and i think it will be different if i take P(x) and Q(x) both linear functions, or any other functions...

Last edited: Mar 6, 2009
4. Mar 7, 2009

### yyat

Re: a weird function i guess....

It looks like you will have to distinguish between different cases. So assume first, for example, that all values are positive, then work out the other cases where different endpoint values are negative.

5. Mar 7, 2009

### songoku

Re: a weird function i guess....

and i also have to work on different assumption of function?

like if i take P(x) and Q(x) linear, then i take P(x) linear and Q(x) exponential

6. Mar 8, 2009

### HallsofIvy

Staff Emeritus
Re: a weird function i guess....

You shouldn't have to do that.

7. Mar 9, 2009

### songoku

Re: a weird function i guess....

but i think it will be different. I and my friends had tried it and we got different answers. I assume that P(x) is linear and Q(x) is exponential and my friend assume that both are linear.

in my opinion, P(x) is increasing doesn't mean that the value of P(x) will be greater than Q(x) which is decreasing function, so we can't say that Q(a) less than P(a).

8. Aug 26, 2009

### songoku

Re: a weird function i guess....

Hi everyone

I think my question hasn't been answered yet so I'm asking for more help. I re-read my question and found out that there is missing information about the question. I'm really sorry about it. Because I can't edit the post, I will re-state the question.

Question :
P(x) is an increasing function and Q(x) is a decreasing function in interval a ≤x≤b, where P(x) and Q(x) lie on the first quadrant. Another function γ(x) satisfies m≤γ(x)≤M.
Find m and M in terms of P(a), P(b), Q(a), and Q(b) if:
a. γ(x) = P(x) . Q(x)
b. γ(x) = [P(x)]^2 – [Q(x)]^2
c. γ(x) = 1/P(X) +Q(x)
d. γ(x)= (P(x))/(Q(x))-(Q(x))/(P(x))

*the hightlight part is the information that I missed before. Sorry :uhh:

So, my work which take the negative and positive value of P(x) and Q(x) into consideration is out of the way because both function are always positive.
But I still think that m and M can't be determined unless we know what kind of function P(x) and Q(x) are. (linear or exponential or others )

Even for both P(x) and Q(x) are linear, the answer still can't be determined. Assume P(a) = 2, P(b) = 4, Q(a) = 4, and Q(b) = 2. Then for γ(x) = P(x) . Q(x) , P(a).Q(a) = 8 and P(b)*Q(b)=8.
The result will be different if we assume P(a) = 1, P(b) = 1000, Q(a) = 4, Q(b) = 2

And I think P(x) and Q(x) can't be zero in that interval because the (c) and (d) questions will have zero denominator.

That's all that I can come up with for now

Thanks a lot and once again, Sorry