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 \(x \geq 4\).