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 \(x > 4\).