Answer :
Sure, let's solve these inequalities step-by-step.
### Part (c) [tex]\( |2x + 1| > -x^2 + 4 \)[/tex]
To solve this inequality, we need to analyze the absolute value function [tex]\( |2x + 1| \)[/tex] and the quadratic [tex]\( -x^2 + 4 \)[/tex]. The inequality asks us to find the values of [tex]\( x \)[/tex] for which the absolute value is greater than the quadratic.
1. Break down the absolute value inequality:
The absolute value inequality [tex]\( |A| > B \)[/tex] can be split into two inequalities: [tex]\( A > B \)[/tex] or [tex]\( -A > B \)[/tex].
[tex]\[ |2x + 1| > -x^2 + 4 \][/tex]
This can be split into:
[tex]\[ (2x + 1) > -x^2 + 4 \quad \text{or} \quad -(2x + 1) > -x^2 + 4 \][/tex]
2. Solving [tex]\( 2x + 1 > -x^2 + 4 \)[/tex]:
[tex]\[ 2x + 1 > -x^2 + 4 \][/tex]
Bringing all terms to one side:
[tex]\[ x^2 + 2x + 1 - 4 > 0 \][/tex]
[tex]\[ x^2 + 2x - 3 > 0 \][/tex]
Factoring the quadratic:
[tex]\[ (x + 3)(x - 1) > 0 \][/tex]
This inequality holds for [tex]\( x < -3 \)[/tex] or [tex]\( x > 1 \)[/tex].
3. Solving [tex]\( -(2x + 1) > -x^2 + 4 \)[/tex]:
[tex]\[ - (2x + 1) > -x^2 + 4 \][/tex]
Simplifying this, we get:
[tex]\[ -2x - 1 > -x^2 + 4 \][/tex]
Bringing all terms to one side:
[tex]\[ x^2 - 2x - 5 < 0 \][/tex]
Notice this does not factor nicely, but we can solve it using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \quad \text{where } a = 1, b = -2, c = -5 \][/tex]
[tex]\[ x = \frac{2 \pm \sqrt{4 + 20}}{2} \][/tex]
[tex]\[ x = 1 \pm \sqrt{6} \][/tex]
So, [tex]\( 1 - \sqrt{6} < x < 1 + \sqrt{6} \)[/tex]
4. Combining the results:
Combining the intervals [tex]\( x < -3 \)[/tex], [tex]\( x > 1 \)[/tex] from [tex]\( (2x + 1 > -x^2 + 4) \)[/tex] and the interval [tex]\( 1 - \sqrt{6} < x < 1 + \sqrt{6} \)[/tex], we get:
[tex]\[ x < 1 - \sqrt{6} \quad \text{or} \quad x > 1 \][/tex]
### Part (d) [tex]\( \left|\frac{x+3}{x-1}\right| \leq 3 \)[/tex]
To solve this inequality, we need to analyze the absolute value function [tex]\( \left|\frac{x+3}{x-1}\right| \leq 3 \)[/tex].
1. Break down the absolute value inequality:
The absolute value inequality [tex]\( |A| \leq B \)[/tex] can be split into two inequalities: [tex]\( -B \leq A \leq B \)[/tex].
[tex]\[ -3 \leq \frac{x+3}{x-1} \leq 3 \][/tex]
2. Solving [tex]\( \frac{x+3}{x-1} \leq 3 \)[/tex]:
Solving:
[tex]\[ \frac{x+3}{x-1} \leq 3 \][/tex]
This can be rearranged to:
[tex]\[ x+3 \leq 3(x-1) \][/tex]
Simplifying:
[tex]\[ x + 3 \leq 3x - 3 \][/tex]
[tex]\[ 6 \leq 2x \][/tex]
[tex]\[ 3 \leq x \][/tex]
3. Solving [tex]\( -3 \leq \frac{x+3}{x-1} \)[/tex]:
Solving:
[tex]\[ -3 \leq \frac{x+3}{x-1} \][/tex]
This can be rearranged to:
[tex]\[ -3(x-1) \leq x+3 \][/tex]
Simplifying:
[tex]\[ -3x + 3 \leq x + 3 \][/tex]
[tex]\[ -3x \leq x \][/tex]
[tex]\[ -3 \leq x \][/tex]
This combined with the previous results we get:
[tex]\[ 3 \leq x \quad \text{or} \quad x \leq 0 \][/tex]
4. Combining the results:
Combining the intervals [tex]\( x \geq 3 \)[/tex] or [tex]\( x \leq 0 \)[/tex], we get:
[tex]\[ x \leq 0 \quad \text{or} \quad x \geq 3 \][/tex]
In conclusion, the solutions to the inequalities are:
[tex]\[ \boxed{\text{(c)}\quad x < 1 - \sqrt{6} \quad \text{or} \quad x > 1} \][/tex]
[tex]\[ \boxed{\text{(d)}\quad x \leq 0 \quad \text{or} \quad x \geq 3} \][/tex]
### Part (c) [tex]\( |2x + 1| > -x^2 + 4 \)[/tex]
To solve this inequality, we need to analyze the absolute value function [tex]\( |2x + 1| \)[/tex] and the quadratic [tex]\( -x^2 + 4 \)[/tex]. The inequality asks us to find the values of [tex]\( x \)[/tex] for which the absolute value is greater than the quadratic.
1. Break down the absolute value inequality:
The absolute value inequality [tex]\( |A| > B \)[/tex] can be split into two inequalities: [tex]\( A > B \)[/tex] or [tex]\( -A > B \)[/tex].
[tex]\[ |2x + 1| > -x^2 + 4 \][/tex]
This can be split into:
[tex]\[ (2x + 1) > -x^2 + 4 \quad \text{or} \quad -(2x + 1) > -x^2 + 4 \][/tex]
2. Solving [tex]\( 2x + 1 > -x^2 + 4 \)[/tex]:
[tex]\[ 2x + 1 > -x^2 + 4 \][/tex]
Bringing all terms to one side:
[tex]\[ x^2 + 2x + 1 - 4 > 0 \][/tex]
[tex]\[ x^2 + 2x - 3 > 0 \][/tex]
Factoring the quadratic:
[tex]\[ (x + 3)(x - 1) > 0 \][/tex]
This inequality holds for [tex]\( x < -3 \)[/tex] or [tex]\( x > 1 \)[/tex].
3. Solving [tex]\( -(2x + 1) > -x^2 + 4 \)[/tex]:
[tex]\[ - (2x + 1) > -x^2 + 4 \][/tex]
Simplifying this, we get:
[tex]\[ -2x - 1 > -x^2 + 4 \][/tex]
Bringing all terms to one side:
[tex]\[ x^2 - 2x - 5 < 0 \][/tex]
Notice this does not factor nicely, but we can solve it using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \quad \text{where } a = 1, b = -2, c = -5 \][/tex]
[tex]\[ x = \frac{2 \pm \sqrt{4 + 20}}{2} \][/tex]
[tex]\[ x = 1 \pm \sqrt{6} \][/tex]
So, [tex]\( 1 - \sqrt{6} < x < 1 + \sqrt{6} \)[/tex]
4. Combining the results:
Combining the intervals [tex]\( x < -3 \)[/tex], [tex]\( x > 1 \)[/tex] from [tex]\( (2x + 1 > -x^2 + 4) \)[/tex] and the interval [tex]\( 1 - \sqrt{6} < x < 1 + \sqrt{6} \)[/tex], we get:
[tex]\[ x < 1 - \sqrt{6} \quad \text{or} \quad x > 1 \][/tex]
### Part (d) [tex]\( \left|\frac{x+3}{x-1}\right| \leq 3 \)[/tex]
To solve this inequality, we need to analyze the absolute value function [tex]\( \left|\frac{x+3}{x-1}\right| \leq 3 \)[/tex].
1. Break down the absolute value inequality:
The absolute value inequality [tex]\( |A| \leq B \)[/tex] can be split into two inequalities: [tex]\( -B \leq A \leq B \)[/tex].
[tex]\[ -3 \leq \frac{x+3}{x-1} \leq 3 \][/tex]
2. Solving [tex]\( \frac{x+3}{x-1} \leq 3 \)[/tex]:
Solving:
[tex]\[ \frac{x+3}{x-1} \leq 3 \][/tex]
This can be rearranged to:
[tex]\[ x+3 \leq 3(x-1) \][/tex]
Simplifying:
[tex]\[ x + 3 \leq 3x - 3 \][/tex]
[tex]\[ 6 \leq 2x \][/tex]
[tex]\[ 3 \leq x \][/tex]
3. Solving [tex]\( -3 \leq \frac{x+3}{x-1} \)[/tex]:
Solving:
[tex]\[ -3 \leq \frac{x+3}{x-1} \][/tex]
This can be rearranged to:
[tex]\[ -3(x-1) \leq x+3 \][/tex]
Simplifying:
[tex]\[ -3x + 3 \leq x + 3 \][/tex]
[tex]\[ -3x \leq x \][/tex]
[tex]\[ -3 \leq x \][/tex]
This combined with the previous results we get:
[tex]\[ 3 \leq x \quad \text{or} \quad x \leq 0 \][/tex]
4. Combining the results:
Combining the intervals [tex]\( x \geq 3 \)[/tex] or [tex]\( x \leq 0 \)[/tex], we get:
[tex]\[ x \leq 0 \quad \text{or} \quad x \geq 3 \][/tex]
In conclusion, the solutions to the inequalities are:
[tex]\[ \boxed{\text{(c)}\quad x < 1 - \sqrt{6} \quad \text{or} \quad x > 1} \][/tex]
[tex]\[ \boxed{\text{(d)}\quad x \leq 0 \quad \text{or} \quad x \geq 3} \][/tex]
