Solve the following inequalities, if it is known that function g is decreasing on its domain.
1. g(3x^2−2x)≥g(3x−2), Dg=ℝ
2. g(x^3−4x)
3. g(5−x^2)≤g(3x−5), Dg=(−∞,4)
4. g(x^2−3x)≥g(4x−12), Dg=(−∞,0]
But what do we do if the domain isn't all real numbers like the 3rd and 4th example?
If g is decreasing, then if g(a) > g(b), then a<b
so, for #1, since g(3x^2−2x)≥g(3x−2),
3x^2-2x <= 3x-2
3x^2 - 5x + 2 <= 0
(3x-2)(x-1) <= 0
2/3 <= x <= 1
Do the others in like wise.
1. Well, if g is decreasing, it means that as the input of g increases, the output decreases. So, we can say that g(3x^2 - 2x) ≥ g(3x - 2) if and only if 3x^2 - 2x ≥ 3x - 2. Now, let's solve this inequality:
3x^2 - 2x ≥ 3x - 2
3x^2 - 5x + 2 ≥ 0
(3x - 2)(x - 1) ≥ 0
This inequality holds true when (3x - 2) and (x - 1) have the same sign. For that, we have two cases:
Case 1: (3x - 2) > 0 and (x - 1) > 0, which gives x > 2/3 and x > 1. Since we need both conditions to hold, the solution is x > 2/3.
Case 2: (3x - 2) < 0 and (x - 1) < 0, which gives x < 2/3 and x < 1. Again, we need both conditions to hold, so the solution is x < 1.
Therefore, the solution to the inequality is x < 1.
2. Similar to the previous case, g(x^3 - 4x) ≥ g(0) if and only if x^3 - 4x ≥ 0. Let's solve this inequality:
x^3 - 4x ≥ 0
x(x^2 - 4) ≥ 0
x(x + 2)(x - 2) ≥ 0
So, the inequality holds true when x ≤ -2 or -2 ≤ x ≤ 2. But since we are given Dg = ℝ, the solution here is x ∈ ℝ.
3. This time we are given Dg = (-∞, 4) and g(5 - x^2) ≤ g(3x - 5). Since g is decreasing, it means that as the input increases, the output decreases. So, let's solve the inequality:
5 - x^2 ≤ 3x - 5
-x^2 - 3x + 10 ≤ 0
This quadratic inequality can be solved by factorizing:
-(x + 5)(x - 2) ≤ 0
This inequality holds true when (x + 5) and (x - 2) have opposite signs. For that, we have two cases:
Case 1: (x + 5) > 0 and (x - 2) < 0, which gives x > -5 and x < 2. But since Dg = (-∞, 4), the solution is -5 < x < 2.
Case 2: (x + 5) < 0 and (x - 2) > 0, which gives x < -5 and x > 2. However, this case doesn't satisfy Dg = (-∞, 4).
Therefore, the solution to the inequality is -5 < x < 2.
4. In this case, we have g(x^2 - 3x) ≥ g(4x - 12), with Dg = (-∞, 0]. Let's solve the inequality:
x^2 - 3x ≥ 4x - 12
x^2 - 7x + 12 ≥ 0
(x - 3)(x - 4) ≥ 0
This inequality holds true when (x - 3) and (x - 4) have the same sign. Since we are given Dg = (-∞, 0], the solution is x ≤ 3.
Therefore, the solution to the inequality is x ≤ 3.
To solve the given inequalities using the fact that function g is decreasing on its domain, we need to compare the values of the expressions inside the function g.
1. For the inequality g(3x^2−2x) ≥ g(3x−2), we can directly compare the expressions inside g because g is decreasing. Therefore, we have 3x^2−2x ≥ 3x−2.
To solve this quadratic inequality, let's move all the terms to one side:
3x^2−2x − (3x−2) ≥ 0
3x^2−2x − 3x + 2 ≥ 0
3x^2 − 5x + 2 ≥ 0
Next, we can factorize the quadratic expression:
(3x−2)(x−1) ≥ 0
To find the values of x that satisfy this inequality, we can create a number line:
---------|-----|---------
2 1
From the number line, we can see that the solution is x ≤ 1 or x ≥ 2/3. Therefore, the solution to the inequality is the interval (-∞, 1] ∪ [2/3, +∞).
2. For the inequality g(x^3−4x), we can again compare the expressions inside g because g is decreasing. Therefore, x^3−4x ≥ 0.
We can factorize the expression:
x(x^2−4) ≥ 0
x(x−2)(x+2) ≥ 0
We create a number line:
-----|---0---|---2---|---(-2)---|----
-2 0 2
From the number line, we can see that the solution is x ≤ -2 or 0 ≤ x ≤ 2. Therefore, the solution to the inequality is the interval (-∞, -2] ∪ [0, 2].
3. For the inequality g(5−x^2) ≤ g(3x−5), we again compare the expressions inside g due to the decrease property. Therefore, 5−x^2 ≤ 3x−5.
Moving all the terms to one side, we have:
-x^2 - 3x + 10 ≤ 0
However, we also need to incorporate the domain of g, which is given as (-∞, 4). Therefore, we need to find the intersection of the solution to the inequality with the interval (-∞, 4).
We can factorize the quadratic expression:
-(x + 5)(x - 2) ≤ 0
We create a number line:
-------|---(-5)---|---2---|-------
-5 2
From the number line, we can see that the solution is -5 ≤ x ≤ 2. However, since we need to intersect this solution with the given interval (-∞, 4), the final solution is: (-5, 2].
4. For the inequality g(x^2−3x) ≥ g(4x−12), again, we compare the expressions inside g because g is decreasing. Therefore, x^2−3x ≥ 4x−12.
Moving all the terms to one side, we have:
x^2 - 7x + 12 ≥ 0
To solve this inequality, we can factorize the quadratic expression:
(x - 3)(x - 4) ≥ 0
We create a number line:
------|----3----|----4----|------
3 4
From the number line, we can see that the solution is x ≤ 3 or x ≥ 4. Therefore, the solution to the inequality is the interval (-∞,3] ∪ [4, +∞).
When the domain is constricted, you'll have all equations set to the domain. Allow me to explain.
For instance "g(5−x^2)≤g(3x−5), Dg=(−∞,4)" (assuming it is increasing) Will give us the equation 5−x^2≤3x−5. To make sure the domain is in check, you got to create two more equations. These equations, as per the domain, are 5−x^2<4 and 3x−5<4 cuz x<4. Put em in a system and you have the answer.