Given the function \(f(x) = x^2 - 6x + c\), we need \(f(x) > 2\) for all \(x\).

Rearrange the inequality:

\(x^2 - 6x + c > 2\)

\(x^2 - 6x + c - 2 > 0\)

Complete the square for \(x^2 - 6x\):

\((x-3)^2 - 9\)

Substitute back:

\((x-3)^2 - 9 + c - 2 > 0\)

\((x-3)^2 + c - 11 > 0\)

Since \((x-3)^2 \geq 0\) for all \(x\), we need:

\(c - 11 > 0\)

Thus, \(c > 11\).

To solve the equation \(3x + 2 = \frac{2}{x - 1}\), follow these steps:

1. Multiply both sides by \(x - 1\) to eliminate the fraction:

\((3x + 2)(x - 1) = 2\)

2. Expand the left side:

\(3x^2 - 3x + 2x - 2 = 2\)

\(3x^2 - x - 2 = 2\)

3. Bring all terms to one side to form a quadratic equation:

\(3x^2 - x - 4 = 0\)

4. Factor the quadratic equation:

\((3x - 4)(x + 1) = 0\)

5. Solve for \(x\) by setting each factor to zero:

\(3x - 4 = 0\) or \(x + 1 = 0\)

\(3x = 4\) gives \(x = \frac{4}{3}\)

\(x = -1\)

Thus, the solutions are \(x = -1\) and \(x = \frac{4}{3}\).

1. Start with the inequality:

\(x^2 - 4x + 8 < 9\)

2. Simplify the inequality:

\(x^2 - 4x + 8 - 9 < 0\)

\(x^2 - 4x - 1 < 0\)

3. Complete the square for \(x^2 - 4x\):

\((x-2)^2 - 4 - 1 < 0\)

\((x-2)^2 < 5\)

4. Solve the inequality:

\(-\sqrt{5} < x - 2 < \sqrt{5}\)

5. Add 2 to each part:

\(2 - \sqrt{5} < x < 2 + \sqrt{5}\)

Start with the inequality:

\(2x^2 - 6x + 5 > 13\)

Simplify the inequality:

\(2x^2 - 6x + 5 - 13 > 0\)

\(2x^2 - 6x - 8 > 0\)

Factor the quadratic:

\(2(x^2 - 3x - 4) > 0\)

\(2(x - 4)(x + 1) > 0\)

Find the critical points by setting the factors to zero:

\(x - 4 = 0 \Rightarrow x = 4\)

\(x + 1 = 0 \Rightarrow x = -1\)

Test intervals around the critical points:

• For \(x < -1\), choose \(x = -2\): \(2(-2 - 4)(-2 + 1) > 0 \Rightarrow 2(-6)(-1) > 0 \Rightarrow 12 > 0\)
• For \(-1 < x < 4\), choose \(x = 0\): \(2(0 - 4)(0 + 1) > 0 \Rightarrow 2(-4)(1) > 0 \Rightarrow -8 > 0\) (false)
• For \(x > 4\), choose \(x = 5\): \(2(5 - 4)(5 + 1) > 0 \Rightarrow 2(1)(6) > 0 \Rightarrow 12 > 0\)

Thus, the solution is \(x > 4\) or \(x < -1\).

Start with the inequality:

\(x^2 + 6x + 2 > 9\)

Simplify it:

\(x^2 + 6x + 2 - 9 > 0\)

\(x^2 + 6x - 7 > 0\)

Factor the quadratic expression:

\((x + 7)(x - 1) > 0\)

The critical points are \(x = -7\) and \(x = 1\).

Test intervals around the critical points:

1. For \(x < -7\), choose \(x = -8\):

\((-8 + 7)(-8 - 1) > 0\)

\((-1)(-9) > 0\)

\(9 > 0\) (True)

2. For \(-7 < x < 1\), choose \(x = 0\):

\((0 + 7)(0 - 1) > 0\)

\(7(-1) > 0\)

\(-7 > 0\) (False)

3. For \(x > 1\), choose \(x = 2\):

\((2 + 7)(2 - 1) > 0\)

\(9(1) > 0\)

\(9 > 0\) (True)

Thus, the solution is \(x > 1\) or \(x < -7\).

Given the function \(f(x) = 6x - x^2 - 5\), we need to find the values of \(x\) such that \(f(x) \leq 3\).

Start by setting up the inequality:

\(6x - x^2 - 5 \leq 3\)

Rearrange the inequality:

\(-x^2 + 6x - 5 \leq 3\)

\(-x^2 + 6x - 8 \leq 0\)

Multiply through by \(-1\) (which reverses the inequality):

\(x^2 - 6x + 8 \geq 0\)

Factor the quadratic:

\((x - 2)(x - 4) \geq 0\)

The critical points are \(x = 2\) and \(x = 4\).

Test intervals around the critical points:

• For \(x < 2\), choose \(x = 0\): \((0 - 2)(0 - 4) = 8 \geq 0\)
• For \(2 < x < 4\), choose \(x = 3\): \((3 - 2)(3 - 4) = -1 \lt 0\)
• For \(x > 4\), choose \(x = 5\): \((5 - 2)(5 - 4) = 3 \geq 0\)

Thus, the solution is \(x < 2\) or \(x \geq 4\).

To solve the inequality \(4x^2 - 12x > 7\), follow these steps:

1. Rearrange the inequality:

\(4x^2 - 12x - 7 > 0\)

2. Factor the quadratic expression:

\(4x^2 - 12x - 7 = (2x - 7)(2x + 1)\)

3. Solve the inequality \((2x - 7)(2x + 1) > 0\):

- The critical points are \(x = \frac{7}{2}\) and \(x = -\frac{1}{2}\).

4. Test intervals around the critical points:

- For \(x < -\frac{1}{2}\), choose \(x = -1\):

\((2(-1) - 7)(2(-1) + 1) = (-9)(-1) = 9 > 0\)

- For \(-\frac{1}{2} < x < \frac{7}{2}\), choose \(x = 0\):

\((2(0) - 7)(2(0) + 1) = (-7)(1) = -7 < 0\)

- For \(x > \frac{7}{2}\), choose \(x = 4\):

\((2(4) - 7)(2(4) + 1) = (1)(9) = 9 > 0\)

5. The solution is:

\(x > \frac{7}{2} \text{ or } x < -\frac{1}{2}\)

To find the set of \(x\) values for which \(y > 9\), start with the inequality:

\(2x^2 - 3x > 9\)

Subtract 9 from both sides:

\(2x^2 - 3x - 9 > 0\)

Factor the quadratic expression:

\((2x + 3)(x - 3) > 0\)

Find the roots by setting each factor to zero:

\(2x + 3 = 0 \Rightarrow x = -\frac{3}{2}\)

\(x - 3 = 0 \Rightarrow x = 3\)

Test intervals around the roots to determine where the inequality holds:

• For \(x < -\frac{3}{2}\), choose \(x = -2\): \((2(-2) + 3)((-2) - 3) = (-1)(-5) = 5 > 0\)
• For \(-\frac{3}{2} < x < 3\), choose \(x = 0\): \((2(0) + 3)(0 - 3) = (3)(-3) = -9 < 0\)
• For \(x > 3\), choose \(x = 4\): \((2(4) + 3)(4 - 3) = (11)(1) = 11 > 0\)

The solution is \(x > 3\) or \(x < -\frac{3}{2}\).

To find the set of values of \(x\) for which \(f(x) > 4\), we start with the inequality:

\(x^2 - 3x > 4\)

Rearrange the inequality:

\(x^2 - 3x - 4 > 0\)

Factor the quadratic expression:

\((x - 4)(x + 1) > 0\)

Determine the critical points by setting each factor to zero:

\(x - 4 = 0 \Rightarrow x = 4\)

\(x + 1 = 0 \Rightarrow x = -1\)

Test intervals around the critical points:

• For \(x < -1\), choose \(x = -2\): \((x - 4)(x + 1) = (-2 - 4)(-2 + 1) = 6 > 0\)
• For \(-1 < x < 4\), choose \(x = 0\): \((x - 4)(x + 1) = (0 - 4)(0 + 1) = -4 < 0\)
• For \(x > 4\), choose \(x = 5\): \((x - 4)(x + 1) = (5 - 4)(5 + 1) = 6 > 0\)

Thus, the solution is \(x < -1\) or \(x > 4\).