Increasing and Decreasing Function

[hongsgirl]

Hi all
I'm new here and I need help on this question

P(x) is an increasing function and Q(x) is a decreasing function in interval \(\displaystyle a\leq x\leq b\), x is positive. Another function \(\displaystyle \gamma (x)\) satisfies \(\displaystyle m\leq \gamma (x)\leq M\). Find the value of m and M if:

a.\(\displaystyle \gamma (x)= P(x) . Q(x)\)

b.\(\displaystyle \gamma (x)= [P(x)]^2 - [Q(x)]^2\)

c.\(\displaystyle \gamma (x) =\frac{1}{P(x)}+Q(x)\)

d. \(\displaystyle \gamma (x) = \frac{P(x)}{Q(x)}-\frac{Q(x)}{P(x)}\)


Attempt :

P(a)>P(b) and Q(a)<Q(b)

I can't solve this problem because I think we need to know that kind of functions P(x) and Q(x) are.

My thoughts :
1. If P(x) is exponential function and Q(x) is linear function, the result will be different compared to P(x) is linear and Q(x) is exponential. It also will be different if both P(x) and Q(x) are linear.

2. 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 \(\displaystyle \gamma (x) = P(x)*Q(x)\) , P(a)*Q(a) = 8 and P(b)*Q(b)=8.
The result will be different if we assume P(a) = 0, P(b) = 1000, Q(a) = 4, Q(b) = 2.

Or there are flaws in my attempts?

Thanks

 

[daon]

edit to note: I used P(x) instead of P in the beginning, but I mean them as their function, not evaluated at x. I was unsure if "increasing" meant monotonic increasing or just increasing, that is why I did not stop the work at P=2. It should also be fairly obvious looking at the assumptions that P and Q are not zero, nor if they are constant can they be equal.

From d) and b)

\(\displaystyle \gamma(x) =^d \frac{P(x)}{Q(x)} - \frac{Q(x)}{P(x)} = \frac{[P(x)]^2 -[Q(x)]^2}{(P\cdot Q)(x)} = ^b \frac{[P(x)]^2 - [Q(x)]^2}{1}\)

From the above together with a) we get

Thus \(\displaystyle \gamma(x) = (P \cdot Q)(x) = 1\)

From here you have \(\displaystyle m \le 1 \le M\).

But, nevermind the M's... I see a problem with c)

With the above we can get that:

\(\displaystyle 1-Q = \frac{1}{P} \implies P - (P\cdot Q) = 1 \implies P-1=1 \implies P=2\)

And, from a) we must then have \(\displaystyle Q = \frac{1}{2}\)

This makes both b) and d) false.

Your set of premises contradict themselves.

 

[hongsgirl]

Hello daon

edit to note: I used P(x) instead of P in the beginning, but I mean them as their function, not evaluated at x.

Sorry, i don't get this part...

I was unsure if "increasing" meant monotonic increasing or just increasing, that is why I did not stop the work at P=2. It should also be fairly obvious looking at the assumptions that P and Q are not zero, nor if they are constant can they be equal.

I think "increasing" means monotonic increasing, although it's not stated in the question. Oh yes I agree that P(x) and Q(x) can't be zero in interval a and b since the denominator for (c) and (d) will be zero.

From d) and b)

\(\displaystyle \gamma(x) =^d \frac{P(x)}{Q(x)} - \frac{Q(x)}{P(x)} = \frac{[P(x)]^2 -[Q(x)]^2}{(P\cdot Q)(x)} = ^b \frac{[P(x)]^2 - [Q(x)]^2}{1}\)

From the above together with a) we get

Thus \(\displaystyle \gamma(x) = (P \cdot Q)(x) = 1\)

I also don't get this part. Why can P(x)*Q(x) = 1 ?? I don't think question (a) related to other sub-question such as (b), (c), etc


Thoroughly, I don't get your idea...Sorry

Thanks

 

[daon]

Sorry, it seemed to me that you were assuming a->d. I didn't realize this was four separate questions. :lol:

 

[hongsgirl]

hahahaha

So, have any ideas, daon?

 

[daon]

What class is this for? Are you supposed to apply any calculus i.e. derivatives etc.? Are the functions continuous? It seems the answer must depend only on P,Q, a and b. There are also many answers as if you find M and m, then M'=(M+1) and m'=(m-1) will also work.

What I can come up with not using any calculus, but there might be a better solution, for (a) that will work in all situations is:

\(\displaystyle M = \max\{|P(a)|,|P(b)|,|Q(a)|,|Q(b)|\}^2\)
\(\displaystyle m = -M\)


(You can replace the square with the product on the "second maximum," the square makes it neater)

Choosing M as above,
\(\displaystyle M \ge |P(x)Q(x)| \ge P(x)Q(x)\)

I was tempted to use \(\displaystyle M = |P(b)Q(a)|\), however if P(a)=-1, P(b)=0, Q(a)=-1, Q(b) =-2 then

\(\displaystyle |P(b)Q(a)| = 0 \not \ge 1 = Q(a)P(a)\)

Hope that helps some.

 

[hongsgirl]

What class is this for? Are you supposed to apply any calculus i.e. derivatives etc.? Are the functions continuous? It seems the answer must depend only on P,Q, a and b. .

This is past paper for A level exam. I'm not sure what branch this problem is.
I suppose the functions continuous and I also think m and M must be stated in terms of P(a), P(b), Q(a), and Q(b).

There are also many answers as if you find M and m, then M'=(M+1) and m'=(m-1) will also work

I'm not sure about this because we restrict the function in interval a and b

What I can come up with not using any calculus, but there might be a better solution, for (a) that will work in all situations is:

\(\displaystyle M = \max\{|P(a)|,|P(b)|,|Q(a)|,|Q(b)|\}^2\)
\(\displaystyle m = -M\)

(You can replace the square with the product on the "second maximum," the square makes it neater)

Sorry, I don't understand this part...

I was tempted to use \(\displaystyle M = |P(b)Q(a)|\), however if P(a)=-1, P(b)=0, Q(a)=-1, Q(b) =-2 then

\(\displaystyle |P(b)Q(a)| = 0 \not \ge 1 = Q(a)P(a)\)

I think we can't use P(b)Q(a) because we have to use the same value of x. If \(\displaystyle \gamma (x) = P(x)*Q(x) , \text{then}\; \gamma (a) = P(a)*Q(a)\)


If you are allowed to use calculus, can you find solution for this problem? I still think this question can't be solved...

Thanks

 

[daon]

The problem here is M and m are constants, thus P(x)*Q(x) must be between a fixed number FOR ALL x. My counter example shows M= |P(a)Q(b)| does not work for x=a, as \(\displaystyle \gamma(a) > M\). It doesn;t matter if the domain is restricted between a and b. M and m are constants restricting \(\displaystyle \gamma\), not x. Thus if \(\displaystyle \gamma\) is bwteen m and M, it is also between (m-1) and (M+1). The problem does not specify to find the smallest M and largest m that work

Have you seen the max function? I was letting M be the largest of the square of the absolute values of the four real numbers P(a), P(b), Q(a), Q(b). It is a real number in this problem and it works. Can you do better with Calculus? I'm not positive, but I'm thinking yes. Knowing how "fast" a function is increasing/decreasing might help.

For example, in the simple case that \(\displaystyle \gamma'(x) \le 0 \,\, \forall x\) we know that \(\displaystyle M=\gamma(a)\) and \(\displaystyle m=\gamma(b)\)

You are given that \(\displaystyle P'(x) \ge 0\) and \(\displaystyle Q'(x) \le 0\) for all \(\displaystyle x \in (a,b)\). Try playing with this, I need to drive 6 hours upstate hah.

 

[hongsgirl]

Hi daon

It doesn;t matter if the domain is restricted between a and b. M and m are constants restricting \(\displaystyle \gamma\), not x. Thus if \(\displaystyle \gamma\) is bwteen m and M, it is also between (m-1) and (M+1). The problem does not specify to find the smallest M and largest m that work

Ok I get it now. So it means that we can't find the exact value of m and M for this problem?

Have you seen the max function? I was letting M be the largest of the square of the absolute values of the four real numbers P(a), P(b), Q(a), Q(b). It is a real number in this problem and it works.

Yes it works. But I don't know letting M be the largest of the square of the absolute values of the four real numbers P(a), P(b), Q(a), Q(b) is related to which question, a, b, c,or d?

For example, in the simple case that \(\displaystyle \gamma'(x) \le 0 \,\, \forall x\) we know that \(\displaystyle M=\gamma(a)\) and \(\displaystyle m=\gamma(b)\)

Yes it's true. But how can we state that \(\displaystyle \gamma'(x) \le 0 \,\, \forall x\) ? Or maybe we have to divide the question to several cases?

You are given that \(\displaystyle P'(x) \ge 0\) and \(\displaystyle Q'(x) \le 0\) for all \(\displaystyle x \in (a,b)\). Try playing with this

I've tried but didn't get anything new. Even though we know \(\displaystyle P'(x) \ge 0\) and \(\displaystyle Q'(x) \le 0\) , it's not enough to determine M and m because I think we have to know that kind of function P(x) and Q(x) are. (parabolic or linear or etc...)
And also I think P'(x) > 0 and Q(x) < 0 is more appropriate because I think it's monotonically increasing and monotonically decreasing

Thanks