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- 4(y^2 - 3y + 2) \geq 0
- 4y^2 + 12y - 8 \geq 0
-4y^2 + 12y + 1 \geq 0
Multiply through by \(-1\) (reversing the inequality):
4y^2 - 12y - 1 \leq 0
Solve the quadratic inequality using the quadratic formula:
y = \frac{-(-12) \pm \sqrt{(-12)^2 - 4 \cdot 4 \cdot (-1)}}{2 \cdot 4} = \frac{12 \pm \sqrt{144 + 16}}{8} = \frac{12 \pm \sqrt{160}}{8}
y = \frac{12 \pm 4\sqrt{10}}{8} = \frac{3 \pm \sqrt{10}}{2}
Thus, the range of \(f(x)\) is:
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