# Increasing and Decreasing Function

• Aug 15th 2009, 08:29 PM
songoku
Increasing and Decreasing Function
P(x) is an increasing function and Q(x) is a decreasing function in interval $a\leq x\leq b$ , x is positive. Another function $\gamma (x)$ satisfies $m\leq \gamma (x)\leq M$. Find the value of m and M if:

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

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

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

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

Attempt :

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

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 $\gamma (x) = P(x)*Q(x)$ , 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?

Thx
• Aug 20th 2009, 10:33 PM
• Aug 21st 2009, 06:45 AM
songoku

Yes, I also posted on that forum. But still the question has not been answered yet :)

Thx
• Aug 22nd 2009, 05:33 AM
mr fantastic
Quote:

Originally Posted by songoku