Let $f(x) = x^2 + bx + 2 - 3x + 8$ for all real numbers $x$. For what real values of $b$ is $-2$ not in the range of the function $f(x)$? Express your answer in interval notation.
+177 \(f(x) = x^2 + bx + 2 - 3x + 8 = x^2 + (b - 3)x + 10\).
Observe that f(x) is a family of upward-facing parabolas because the leading coefficient, 1, is positive. This means that, in order for -2 to be not in the range, \(\forall x \in \mathbb{R} : f(x) > -2\) must be true.
\(f(x) > -2 \\ x^2 + (b - 3)x + 10 > -2 \\ x^2 + (b - 3)x + 12 > 0\)
The discriminant is valuable here. The discriminant will indicate how many solutions there are to the corresponding equation \(x^2 + (b - 3)x + 12 = 0\) If the discriminant is less than zero, then -2 is not in the range.
\(b^2 - 4ac = (b - 3)^2 - 4 * 1 * 12 \\ b^2 - 6b + 9 - 48 < 0 \\ b^2 - 6b - 39 < 0\)
Now, use the quadratic formula to find the roots of the corresponding equation.
\(b_{1, 2} = \frac{6 \pm \sqrt{36 - 4 * 1 * -39}}{2} \\ b_{1, 2} = \frac{6 \pm \sqrt{192}}{2} = 3 \pm 4\sqrt{3} \\ b_1 \approx -3.928, b_2 \approx 9.928\)
Since this is an upward-facing parabola, b < 0 when \(3 - 4\sqrt{3} < b < 3 + 4\sqrt{3}\), which corresponds to the values of b when -2 is not in the range. Written in interval notation, \((3 - 4\sqrt{3}, 3 + 4\sqrt{3})\).
