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## 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 1

^{st}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 = P

^{2}(a) - Q

^{2}(a) and M = P

^{2}(b) - Q

^{2}(b). Is this correct?

d. γ'(x) = -P'(x) / P

^{2}(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) = [tex]\frac{P'(x).Q(x)-P(x).Q'(x)}{Q^{2}(x)} - \frac{Q'(x).P(x)-P'(x).Q(x)}{P^{2}}[/tex]. 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.