Solve the following inequalities, if it is known that function f is increasing on its domain.
1. f(x^3−4x)≥f(3x^2+6x), Df=ℝ
2. f(x^4−x) 3. f(4x−3)≥f(2−x^2), Df=(−8,4)
4. f(x^2−5x−7)≤f(5−6x), Df=[−1,∞)
For the first one, because the function is increasing and the domain is all real numbers, you can just get rid of the function sign. this changes it to:
x^3-4x>=3x^2+6x
x^3-3x^2-10x>=0
factor:
x(x-5)(x+2)>=0
use the snake method to solve and you will get:
x belongs from [-2,0]U[5,infinity)
for the others do the same thing, except if you have f(a)>f(b) and the domain is Df>c then you have to make sure that a>c and b>c, other than that though you will solve it the same way as the first.
1. Well, since we know that f is increasing on its domain, we can conclude that f(x^3 - 4x) ≥ f(3x^2 + 6x) for all x in ℝ. So, the inequality is true for all values of x.
2. Ah, this one requires a bit more analysis. Since f is increasing, we can compare the values inside the function to determine the inequalities. We find that x^4 - x ≥ 3 when x ≤ -1 and x^4 - x ≤ 3 when x ≥ 1. So, the inequality holds for x in (-∞, -1] and [1, +∞).
3. Now we're getting into some interesting territory! Since f is increasing, we can compare the values of 4x - 3 and 2 - x^2. We find that 4x - 3 ≥ 2 - x^2 when x ≤ -1 and 4x - 3 ≤ 2 - x^2 when -1 ≤ x ≤ 2. So, the inequality holds for x in (-∞, -1] and [-1, 2].
4. Here we go! We can once again use the fact that f is increasing to compare the values of x^2 - 5x - 7 and 5 - 6x. Turns out that x^2 - 5x - 7 ≤ 5 - 6x for all x in [-1, ∞). So, the inequality holds for x in [-1, ∞).
Hope these solutions brought a smile to your face!
To solve the given inequalities, we need to compare the values of the functions on either side of the inequality sign. Since it is known that function f is increasing on its domain, we can directly compare the expressions inside the function.
1. f(x^3 - 4x) ≥ f(3x^2 + 6x)
In this case, we have two expressions inside the function f. Both expressions involve polynomials, and since f is increasing, we can compare the polynomials directly.
Comparing the two expressions, we have:
x^3 - 4x ≥ 3x^2 + 6x
Rearranging and simplifying the inequality:
x^3 - 4x - 3x^2 - 6x ≥ 0
x^3 - 3x^2 - 10x ≥ 0
We can factor out an x from the left side:
x(x^2 - 3x - 10) ≥ 0
Now, we need to find the values of x that satisfy this inequality. We can solve it by factoring the quadratic expression:
(x - 5)(x + 2)(x - 1) ≥ 0
The critical points are x = -2, 1, and 5. We can divide the number line into four intervals: (-∞, -2), (-2, 1), (1, 5), and (5, ∞).
Now, we test each interval by picking a test point from each interval and substituting it into the inequality.
For (-∞, -2), we test x = -3:
(-3 - 5)(-3 + 2)(-3 - 1) ≥ 0
(-8)(-1)(-4) ≥ 0
32 ≥ 0 (true)
For (-2, 1), we test x = 0:
(0 - 5)(0 + 2)(0 - 1) ≥ 0
(-5)(2)(-1) ≥ 0
10 ≥ 0 (true)
For (1, 5), we test x = 2:
(2 - 5)(2 + 2)(2 - 1) ≥ 0
(-3)(4)(1) ≥ 0
-12 ≥ 0 (false)
For (5, ∞), we test x = 6:
(6 - 5)(6 + 2)(6 - 1) ≥ 0
(1)(8)(5) ≥ 0
40 ≥ 0 (true)
Based on the test points, we can conclude that the inequality is true for x in (-∞, -2] ∪ [1,5].
2. f(x^4 - x) < 3
In this case, we are given a strict inequality. We need to compare the expression inside the function f with the constant value of 3.
Comparing the two expressions, we have:
x^4 - x < 3
Rearranging and simplifying the inequality:
x^4 - x - 3 < 0
We need to find the values of x that satisfy this inequality. Since the inequality involves a polynomial, we can use various methods such as factoring or the Rational Root Theorem to find the roots. However, since the exact roots may not be easy to find, we will use a graphical method.
Plotting the graph of the equation y = x^4 - x - 3, we can determine the intervals where the function is below the line y = 0.
The graph shows that the function is below the x-axis in the interval approximately (-1.7, 1.7).
Therefore, the solution to the inequality is -1.7 < x < 1.7.
3. f(4x - 3) ≥ f(2 - x^2), Df = (-8, 4)
In this case, we are given two expressions inside the function f involving polynomials. We need to compare the two expressions.
Comparing the two expressions, we have:
4x - 3 ≥ 2 - x^2
Rearranging and simplifying the inequality:
x^2 + 4x - 5 ≥ 0
We can factor the quadratic expression:
(x + 5)(x - 1) ≥ 0
The critical points are x = -5 and x = 1. We divide the number line into three intervals: (-∞, -5), (-5, 1), and (1, ∞).
Now, we test each interval by picking a test point from each interval and substituting it into the inequality.
For (-∞, -5), we test x = -6:
(-6 + 5)(-6 - 1) ≥ 0
(-1)(-7) ≥ 0
7 ≥ 0 (true)
For (-5, 1), we test x = 0:
(0 + 5)(0 - 1) ≥ 0
(5)(-1) ≥ 0
-5 ≥ 0 (false)
For (1, ∞), we test x = 2:
(2 + 5)(2 - 1) ≥ 0
(7)(1) ≥ 0
7 ≥ 0 (true)
Based on the test points, we can conclude that the inequality is true for x in (-∞, -5] ∪ (1, ∞).
4. f(x^2 - 5x - 7) ≤ f(5 - 6x), Df = [-1, ∞)
In this case, we have two expressions inside the function f involving polynomials. We need to compare the two expressions.
Comparing the two expressions, we have:
x^2 - 5x - 7 ≤ 5 - 6x
Rearranging and simplifying the inequality:
x^2 + x - 12 ≤ 0
We can factor the quadratic expression:
(x - 3)(x + 4) ≤ 0
The critical points are x = -4 and x = 3. We divide the number line into three intervals: (-∞, -4), (-4, 3), and (3, ∞).
Now, we test each interval by picking a test point from each interval and substituting it into the inequality.
For (-∞, -4), we test x = -5:
(-5 - 3)(-5 + 4) ≤ 0
(-8)(-1) ≤ 0
8 ≤ 0 (false)
For (-4, 3), we test x = 0:
(0 - 3)(0 + 4) ≤ 0
(-3)(4) ≤ 0
-12 ≤ 0 (true)
For (3, ∞), we test x = 4:
(4 - 3)(4 + 4) ≤ 0
(1)(8) ≤ 0
8 ≤ 0 (false)
Based on the test points, we can conclude that the inequality is true for x in (-4, 3].