Problem #369
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369
The function \(f\) is defined by \(f : x \mapsto 6x - x^2 - 5\) for \(x \in \mathbb{R}\). Find the set of values of \(x\) for which \(f(x) \leq 3\).
Solution
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 \(x \geq 4\).