Range Interval 3B9473

1. **State the problem:** We have a function $$f(x) = x^2 + x - 4$$ defined for rational numbers $$x$$ in the interval $$[a, a+3]$$. The range of $$f$$ on this interval is given as $$-2 \leq f(x) \leq 16$$. We need to find the possible values of $$a$$. 2. **Recall the properties of quadratic functions:** The function $$f(x) = x^2 + x - 4$$ is a quadratic with leading coefficient $$1 > 0$$, so it opens upwards. The vertex of $$f$$ is at $$x = -\frac{b}{2a} = -\frac{1}{2}$$. 3. **Evaluate the vertex value:** $$f\left(-\frac{1}{2}\right) = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) - 4 = \frac{1}{4} - \frac{1}{2} - 4 = -\frac{9}{4} = -2.25$$. 4. **Analyze the interval $$[a, a+3]$$ and range constraints:** Since the parabola opens upward and the vertex is at $$x = -0.5$$ with minimum value $$-2.25$$, the minimum value of $$f(x)$$ on any interval depends on whether the vertex lies inside that interval. 5. **Check if vertex is inside the interval:** The vertex $$-0.5$$ lies inside $$[a, a+3]$$ if and only if $$a \leq -0.5 \leq a+3$$. This implies $$a \leq -0.5$$ and $$a \geq -3.5$$. So, if $$a \in [-3.5, -0.5]$$, the minimum of $$f(x)$$ on $$[a, a+3]$$ is $$f(-0.5) = -2.25$$. 6. **Compare minimum value with range lower bound:** The problem states $$-2 \leq f(x)$$, but the minimum at vertex is $$-2.25 < -2$$. So if the vertex is inside the interval, the minimum value is less than the allowed range. 7. **If vertex is outside the interval:** - If $$a > -0.5$$, then the interval $$[a, a+3]$$ lies to the right of the vertex. - If $$a < -3.5$$, then the interval lies to the left of the vertex. 8. **Calculate $$f(a)$$ and $$f(a+3)$$:** $$f(a) = a^2 + a - 4$$ $$f(a+3) = (a+3)^2 + (a+3) - 4 = a^2 + 6a + 9 + a + 3 - 4 = a^2 + 7a + 8$$ 9. **Since the parabola opens upward, if vertex is outside the interval, the minimum is at one endpoint and maximum at the other.** 10. **Set the range constraints:** $$-2 \leq \min(f(a), f(a+3))$$ and $$\max(f(a), f(a+3)) \leq 16$$ 11. **Analyze cases:** - Case 1: $$a > -0.5$$ (interval right of vertex) - Minimum at $$f(a)$$ - Maximum at $$f(a+3)$$ - Constraints: $$-2 \leq f(a) = a^2 + a - 4$$ $$f(a+3) = a^2 + 7a + 8 \leq 16$$ - Case 2: $$a < -3.5$$ (interval left of vertex) - Minimum at $$f(a+3)$$ - Maximum at $$f(a)$$ - Constraints: $$-2 \leq f(a+3) = a^2 + 7a + 8$$ $$f(a) = a^2 + a - 4 \leq 16$$ 12. **Solve Case 1 inequalities:** - $$-2 \leq a^2 + a - 4 \Rightarrow a^2 + a - 2 \geq 0$$ - $$a^2 + 7a + 8 \leq 16 \Rightarrow a^2 + 7a - 8 \leq 0$$ 13. **Factor and solve:** - $$a^2 + a - 2 = (a+2)(a-1) \geq 0$$ - Solution: $$a \leq -2$$ or $$a \geq 1$$ - $$a^2 + 7a - 8 = (a+8)(a-1) \leq 0$$ - Solution: $$-8 \leq a \leq 1$$ 14. **Combine Case 1 solutions:** $$a \leq -2$$ or $$a \geq 1$$ from first inequality and $$-8 \leq a \leq 1$$ from second Intersection: - For $$a \leq -2$$ and $$-8 \leq a \leq 1$$, intersection is $$-8 \leq a \leq -2$$ - For $$a \geq 1$$ and $$-8 \leq a \leq 1$$, intersection is $$a = 1$$ Also, Case 1 requires $$a > -0.5$$, so $$-8 \leq a \leq -2$$ is invalid here. Only $$a = 1$$ satisfies $$a > -0.5$$. 15. **Solve Case 2 inequalities:** - $$-2 \leq a^2 + 7a + 8 \Rightarrow a^2 + 7a + 10 \geq 0$$ - $$a^2 + a - 4 \leq 16 \Rightarrow a^2 + a - 20 \leq 0$$ 16. **Factor and solve:** - $$a^2 + 7a + 10 = (a+5)(a+2) \geq 0$$ - Solution: $$a \leq -5$$ or $$a \geq -2$$ - $$a^2 + a - 20 = (a+5)(a-4) \leq 0$$ - Solution: $$-5 \leq a \leq 4$$ 17. **Combine Case 2 solutions:** $$a \leq -5$$ or $$a \geq -2$$ from first inequality and $$-5 \leq a \leq 4$$ from second Intersection: - For $$a \leq -5$$ and $$-5 \leq a \leq 4$$, intersection is $$a = -5$$ - For $$a \geq -2$$ and $$-5 \leq a \leq 4$$, intersection is $$-2 \leq a \leq 4$$ Case 2 requires $$a < -3.5$$, so valid values are $$a = -5$$ (since $$-5 < -3.5$$) and any $$a$$ in $$[-2,4]$$ is invalid here. 18. **Check vertex inside interval case:** If $$a \in [-3.5, -0.5]$$, minimum is $$-2.25 < -2$$, so no solution. 19. **Summary of possible $$a$$ values:** - From Case 1: $$a = 1$$ - From Case 2: $$a = -5$$ 20. **Verify range for $$a=1$$:** Interval: $$[1,4]$$ $$f(1) = 1 + 1 - 4 = -2$$ $$f(4) = 16 + 4 - 4 = 16$$ Range: $$[-2,16]$$ matches given range. 21. **Verify range for $$a=-5$$:** Interval: $$[-5, -2]$$ $$f(-5) = 25 - 5 - 4 = 16$$ $$f(-2) = 4 - 2 - 4 = -2$$ Range: $$[-2,16]$$ matches given range. **Final answer:** $$\boxed{a = -5 \text{ or } a = 1}$$