Part 1: The expressions inside the absolute values are \( 4 - 2x \) and \( x - 3 \). We can set each expression equal to zero to find the critical points:
\[ 4 - 2x = 0 \implies x = 2 \]
\[ x - 3 = 0 \implies x = 3 \]
These critical points divide the number line into three intervals:
Interval 1: \( x < 2 \)
In this interval, \( 4 - 2x > 0 \) and \( x - 3 < 0 \). Thus, the inequality becomes:
\[ 4 - 2x \leq -(x - 3) \implies 4 - 2x \leq -x + 3 \]
Solve for \( x \):
\[ 4 - 3 \leq -x + 2x \implies 1 \leq x \]
Since \( x < 2 \), the solution in this interval is \( 1 \leq x < 2 \).
Interval 2: \( 2 \leq x < 3 \)
In this interval, \( 4 - 2x \leq 0 \) and \( x - 3 < 0 \). Thus, the inequality becomes:
\[ -(4 - 2x) \leq -(x - 3) \implies -4 + 2x \leq -x + 3 \]
Solve for \( x \):
\[ 2x + x \leq 3 + 4 \implies 3x \leq 7 \implies x \leq \frac{7}{3} \]
Since \( 2 \leq x < 3 \), the solution in this interval is \( 2 \leq x \leq \frac{7}{3} \).
Interval 3: \( x \geq 3 \)
In this interval, \( 4 - 2x \leq 0 \) and \( x - 3 \geq 0 \). Thus, the inequality becomes:
\[ -(4 - 2x) \leq x - 3 \implies -4 + 2x \leq x - 3 \]
Solve for \( x \):
\[ 2x - x \leq -3 + 4 \implies x \leq 1 \]
Since \( x \geq 3 \), there is no solution in this interval.
Combining the solutions from the intervals, we get:
\[ 1 \leq x \leq \frac{7}{3} \]
CloseTo solve the inequalit:y
\[ \left| \frac{x + 5}{x - 5} \right| \leq 3 \]
Consider the definition of absolute value, which leads to two cases:
Case 1:
\[ \frac{x + 5}{x - 5} \leq 3 \]
\[ \frac{x + 5}{x - 5} - 3 \leq 0 \]
\[ \frac{x + 5 - 3(x - 5)}{x - 5} \leq 0 \]
\[ \frac{x + 5 - 3x + 15}{x - 5} \leq 0 \]
\[ \frac{-2x + 20}{x - 5} \leq 0 \]
\[ \frac{-2(x - 10)}{x - 5} \leq 0 \]
Multiply both sides by \( -1 \) (remembering to reverse the inequality sign):
\[ \frac{2(x - 10)}{x - 5} \geq 0 \]
The critical points are \( x = 5 \) and \( x = 10 \). Testing intervals:
Thus, the solution for Case 1 is \( x < 5 \) or \( x \geq 10 \).
Case 2:
\[ \frac{x + 5}{x - 5} \geq -3 \]
\[ \frac{x + 5}{x - 5} + 3 \geq 0 \]
\[ \frac{x + 5 + 3(x - 5)}{x - 5} \geq 0 \]
\[ \frac{x + 5 + 3x - 15}{x - 5} \geq 0 \]
\[ \frac{4x - 10}{x - 5} \geq 0 \]
\[ \frac{2(2x - 5)}{x - 5} \geq 0 \]
The critical points are \( x = 2.5 \) and \( x = 5 \). Testing intervals:
Thus, the solution for Case 2 is \( x \leq 2.5 \) or \( x > 5 \).
The overall solution is the intersection of the solutions from both cases:
\[ x \leq 2.5 \quad \text{or} \quad x \geq 10 \]
Therefore, the solution to the inequality is:
\[ x \in (-\infty, 2.5] \cup [10, \infty) \]
ClosePart 1: To solve the inequality
\[ |x - 3| - |x - 8| \leq 5 \]
we analyze the expression by considering different intervals based on the critical points where the expressions inside the absolute values change behavior. The critical points are \(x = 3\) and \(x = 8\). We divide the real number line into three intervals:
We solve the inequality in each interval separately:
Case 1: \( x < 3 \)
In this interval, both \(x - 3\) and \(x - 8\) are negative. Therefore:
\[ |x - 3| = -(x - 3) = -x + 3 \]
\[ |x - 8| = -(x - 8) = -x + 8 \]
Substituting into the inequality:
\[ (-x + 3) - (-x + 8) \leq 5 \]
\[ - x + 3 + x - 8 \leq 5 \]
\[ -5 \leq 5 \]
This simplifies to \( -5 \leq 5 \), which is always true. Therefore, all \(x < 3\) satisfy the inequality.
Case 2: \( 3 \leq x < 8 \)
In this interval, \(x - 3\) is non-negative, but \(x - 8\) is negative. Therefore:
\[ |x - 3| = x - 3 \]
\[ |x - 8| = -(x - 8) = -x + 8 \]
Substituting into the inequality:
\[ (x - 3) - (-x + 8) \leq 5 \]
\[ x - 3 + x - 8 \leq 5 \]
\[ 2x - 11 \leq 5 \]
\[ 2x \leq 16 \]
\[ x \leq 8 \]
Since we are already in the interval \( 3 \leq x < 8 \), this condition is satisfied for all \( x \) in this interval.
Case 3: \( x \geq 8 \)
In this interval, both \(x - 3\) and \(x - 8\) are non-negative. Therefore:
\[ |x - 3| = x - 3 \]
\[ |x - 8| = x - 8 \]
Substituting into the inequality:
\[ (x - 3) - (x - 8) \leq 5 \]
\[ x - 3 - x + 8 \leq 5 \]
\[ 5 \leq 5 \]
This simplifies to \( 5 \leq 5 \), which is always true. Therefore, all \(x \geq 8\) satisfy the inequality.
Combining all cases:
The solution to the inequality is the union of all intervals where the inequality holds:
\[ x \in (-\infty, 8] \cup [8, \infty) \]
However, since \(x \geq 8\) is already included in the first interval, the final solution is:
\[ x \in (-\infty, \infty) \]
This means that the inequality holds for all real numbers \(x\).
ClosePart 1: First, rewrite the inequality in standard form:
\[ x^2 - 3x - 4 \geq 0 \]
Factor the quadratic expression:
\[ (x - 4)(x + 1) \geq 0 \]
The critical points are \( x = -1 \) and \( x = 4 \). These divide the number line into three intervals:
Test each interval:
1. For \( x < -1 \): Choose \( x = -2 \):
\[ (-2 - 4)(-2 + 1) = (-6)(-1) = 6 \geq 0 \]
The inequality holds.
2. For \( -1 \leq x \leq 4 \): Choose \( x = 0 \):
\[ (0 - 4)(0 + 1) = (-4)(1) = -4 < 0 \]
The inequality does not hold.
3. For \( x > 4 \): Choose \( x = 5 \):
\[ (5 - 4)(5 + 1) = (1)(6) = 6 \geq 0 \]
The inequality holds.
Therefore, the solution to the inequality is:
\[ x \in (-\infty, -1] \cup [4, \infty) \]
Part 2: Prove True for \( n = 4 \):
\[ 3^4 = 81 \quad \text{and} \quad 4^3 - 3 = 64 - 3 = 61 \]
\[ 81 > 61 \quad \text{is true.} \]
Assume that the statement holds for some \( k \geq 4 \), i.e.,
\[ 3^k > k^3 - 3 \]
We need to show that the statement holds for \( k + 1 \), i.e.,
\[ 3^{k+1} > (k + 1)^3 - 3 \]
Starting from the inductive hypothesis:
\[ 3^{k+1} = 3 \cdot 3^k > 3(k^3 - 3) = 3k^3 - 9 \]
Now, compare \( 3k^3 - 9 \) with \( (k + 1)^3 - 3 \):
\[ (k + 1)^3 - 3 = k^3 + 3k^2 + 3k + 1 - 3 = k^3 + 3k^2 + 3k - 2 \]
We need to show that:
\[ 3k^3 - 9 > k^3 + 3k^2 + 3k - 2 \]
Simplify:
\[ 2k^3 - 3k^2 - 3k - 7 > 0 \]
For \( k \geq 4 \), this inequality holds because the dominant term \( 2k^3 \) grows faster than the other terms. Therefore, the inductive step is valid.
By the principle of mathematical induction, \( 3^n > n^3 - 3 \) for all \( n \in \boldsymbol{{Z}^+} \), \( n \geq 4 \).
Closea) First, factor the denominator:
\[ x^2 - 3x - 4 = (x - 4)(x + 1) \]
Express the function as partial fractions:
\[ \frac{8}{(x - 4)(x + 1)} = \frac{A}{x - 4} + \frac{B}{x + 1} \]
Multiply both sides by the denominator:
\[ 8 = A(x + 1) + B(x - 4) \]
Now, solve for \( A \) and \( B \):
1. Let \( x = 4 \):
\[ 8 = A(4 + 1) + B(4 - 4) \implies 8 = 5A \implies A = \frac{8}{5} \]
2. Let \( x = -1 \):
\[ 8 = A(-1 + 1) + B(-1 - 4) \implies 8 = -5B \implies B = -\frac{8}{5} \]
Therefore, the partial fractions are:
\[ f(x) = \frac{8/5}{x - 4} - \frac{8/5}{x + 1} \]
b) First, factor the denominator:
\[ x^2 - 3x - 4 = (x - 4)(x + 1) \]
Express the function as partial fractions:
\[ \frac{4x + 8}{(x - 4)(x + 1)} = \frac{A}{x - 4} + \frac{B}{x + 1} \]
Multiply both sides by the denominator:
\[ 4x + 8 = A(x + 1) + B(x - 4) \]
Now, solve for \( A \) and \( B \):
1. Let \( x = 4 \):
\[ 4(4) + 8 = A(4 + 1) + B(4 - 4) \implies 24 = 5A \implies A = \frac{24}{5} \]
2. Let \( x = -1 \):
\[ 4(-1) + 8 = A(-1 + 1) + B(-1 - 4) \implies 4 = -5B \implies B = -\frac{4}{5} \]
Therefore, the partial fractions are:
\[ f(x) = \frac{24/5}{x - 4} - \frac{4/5}{x + 1} \]
c) First, factor the denominator:
\[ 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x - 3)(x + 1) \]
Express the function as partial fractions:
\[ \frac{2x}{3(x - 3)(x + 1)} = \frac{A}{x - 3} + \frac{B}{x + 1} \]
Multiply both sides by the denominator:
\[ 2x = 3A(x + 1) + 3B(x - 3) \]
Now, solve for \( A \) and \( B \):
1. Let \( x = 3 \):
\[ 2(3) = 3A(3 + 1) + 3B(3 - 3) \implies 6 = 12A \implies A = \frac{1}{2} \]
2. Let \( x = -1 \):
\[ 2(-1) = 3A(-1 + 1) + 3B(-1 - 3) \implies -2 = -12B \implies B = \frac{1}{6} \]
Therefore, the partial fractions are:
\[ f(x) = \frac{1/2}{x - 3} + \frac{1/6}{x + 1} \]
Close1. Bring each term to one side, so that 0 remains on the right:
\[ x + \frac{3}{x} - 4 \geq 0 \]
2. Combine the terms into a single rational expression:
\[ \frac{x^2 + 3 - 4x}{x} \geq 0 \]
Simplify the numerator:
\[ \frac{x^2 - 4x + 3}{x} \geq 0 \]
3. Factor the numerator:
\[ \frac{(x - 1)(x - 3)}{x} \geq 0 \]
4. Identify the critical points:
The critical points are \( x = 0 \), \( x = 1 \), and \( x = 3 \). These divide the number line into four intervals:
5. Test each interval:
- For \( x < 0 \):
Choice \( x = -1 \):
\[ \frac{(-1 - 1)(-1 - 3)}{-1} = \frac{(-2)(-4)}{-1} = \frac{8}{-1} = -8 < 0 \]
The inequality does not hold.
- For \( 0 < x < 1 \):
Choice \( x = \frac{1}{2} \):
\[ \frac{(\frac{1}{2} - 1)(\frac{1}{2} - 3)}{\frac{1}{2}} = \frac{(-\frac{1}{2})(-\frac{5}{2})}{\frac{1}{2}} = \frac{\frac{5}{4}}{\frac{1}{2}} = \frac{5}{2} > 0 \]
The inequality holds.
- For \( 1 \leq x < 3 \):
Choice \( x = 2 \):
\[ \frac{(2 - 1)(2 - 3)}{2} = \frac{(1)(-1)}{2} = -\frac{1}{2} < 0 \]
The inequality does not hold.
- For \( x \geq 3 \):
Choice \( x = 4 \):
\[ \frac{(4 - 1)(4 - 3)}{4} = \frac{(3)(1)}{4} = \frac{3}{4} > 0 \]
The inequality holds.
6. Consider the critical points:
- At \( x = 0 \), the expression is undefined.
- At \( x = 1 \) and \( x = 3 \), the expression equals zero, which satisfies the inequality.
7. Combine the intervals where the inequality holds:
The solution is \( 0 < x \leq 1 \) or \( x \geq 3 \).
Therefore, the solution to the inequality is:
\[ x \in (0, 1] \cup [3, \infty) \]
Close1. Horizontal Asymptotes:
For rational functions, the horizontal asymptote depends on the degrees of the numerator and the denominator:
In this case, both the numerator and the denominator are of degree 2. Therefore, the horizontal asymptote is:
\[ y = \frac{4}{3} \]
2. Vertical Asymptotes:
Vertical asymptotes occur where the denominator is zero and the numerator is not zero at those points.
First, factor the denominator:
\[ 3x^2 - 9x + 6 = 3(x^2 - 3x + 2) = 3(x - 1)(x - 2) \]
Set the denominator equal to zero to find the vertical asymptotes:
\[ 3(x - 1)(x - 2) = 0 \implies x = 1 \text{ or } x = 2 \]
Check if the numerator is zero at these points:
Therefore, the vertical asymptotes are at \( x = 1 \) and \( x = 2 \).
3. Oblique Asymptotes:
Oblique asymptotes occur when the degree of the numerator is exactly one more than the degree of the denominator. In this case, both the numerator and the denominator are of degree 2, so there is no oblique asymptote.
Close1. Find \((f \circ g)(x)\):
\[ (f \circ g)(x) = f(g(x)) = f\left(\frac{2x}{x - 1}\right) = 3\left(\frac{2x}{x - 1}\right) - 2 = \frac{6x}{x - 1} - 2 \]
Combine the terms:
\[ \frac{6x}{x - 1} - 2 = \frac{6x - 2(x - 1)}{x - 1} = \frac{6x - 2x + 2}{x - 1} = \frac{4x + 2}{x - 1} \]
2. Find \((g \circ f)(x)\):
\[ (g \circ f)(x) = g(f(x)) = g(3x - 2) = \frac{2(3x - 2)}{(3x - 2) - 1} = \frac{6x - 4}{3x - 3} \]
Simplify the denominator:
\[ \frac{6x - 4}{3x - 3} = \frac{6x - 4}{3(x - 1)} = \frac{2(3x - 2)}{3(x - 1)} \]
Rewrite the inequality as:
\[ \frac{4x + 2}{x - 1} - \frac{6x - 4}{3(x - 1)} \leq 0 \]
Combine the fractions:
\[ \frac{3(4x + 2) - (6x - 4)}{3(x - 1)} \leq 0 \]
Simplify the numerator:
\[ \frac{12x + 6 - 6x + 4}{3(x - 1)} \leq 0 \]
\[ \frac{6x + 10}{3(x - 1)} \leq 0 \]
Factor out common terms:
\[ \frac{2(3x + 5)}{3(x - 1)} \leq 0 \]
The critical points are:
These divide the number line into three intervals:
Test the intervals:
At \( x = -\frac{5}{3} \), the expression equals zero.
At \( x = 1 \), the expression is undefined.
The inequality holds when the expression is negative or zero, which occurs in the interval:
\[ x \in \left[-\frac{5}{3}, 1\right) \]
Close