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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.

Aug 11, 2023

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$$f(x) = x^2 + bx + 2 - 3x + 8 = x^2 + (b - 3)x + 10$$.

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, $$(3 - 4\sqrt{3}, 3 + 4\sqrt{3})$$.

Aug 11, 2023