Question

Solve each inequality given that the function f is increasing over its domain

1.f(2x+3)>f(3x-2), Df=[3,infinity)
2f(x^2-2)

Answer 1

Since f is increasing over its domain, we know that if a > b, then f(a) > f(b). Therefore, we have:

2x + 3 > 3x - 2

Solving for x, we get:

x < 5

So the solution set is:

Df = [3, infinity) intersect (-infinity, 5) = [3, 5)

Therefore, the inequality is true for all x in the interval [3, 5).

Since f is increasing over its domain, we know that if a > b, then f(a) > f(b). Therefore, we have:

x^2 - 2 < y^2 - 2

Simplifying, we get:

x^2 < y^2

Taking the square root of both sides, we get:

|x| < |y|

So the solution set is:

Df = [3, infinity)

|x| < |y| means that either x < y or -x < y. Therefore, the solution set can be divided into two parts:

Part 1: x^2 - 2 < 0, i.e., x is in the interval (-sqrt(2), sqrt(2)). For this part, we have:

Df intersect (-sqrt(2), sqrt(2)) = empty set

Part 2: x^2 - 2 >= 0, i.e., x is outside the interval (-sqrt(2), sqrt(2)). For this part, we have:

Df intersect (-infinity, -sqrt(2)] union [sqrt(2), infinity) = [3, infinity)

Therefore, the inequality is true for all x in the interval [3, infinity) except for the interval (-sqrt(2), sqrt(2)).

Explanation: