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Difficulty: HardInequalities and Absolute Value Equations

If xx is a real number satisfying the equation ∣x+3∣+∣x−5∣=8|x + 3| + |x - 5| = 8, what is the maximum possible value of ∣x2−4x−12∣|x^2 - 4x - 12|?

  1. A
    7
  2. B
    9
  3. C
    12
  4. 16Answer
  5. E
    25

Answer

16
The equation ∣x+3∣+∣x−5∣=8|x + 3| + |x - 5| = 8 represents the sum of the distances from xx to −3-3 and from xx to 55. Since the distance between −3-3 and 55 is exactly 88, this equation holds for all xx in the interval [−3,5][-3, 5]. Rewriting x2−4x−12x^2 - 4x - 12 as (x−2)2−16(x - 2)^2 - 16, the vertex occurs at x=2x = 2, which lies inside [−3,5][-3, 5]. At x=2x = 2, x2−4x−12=−16x^2 - 4x - 12 = -16, so ∣x2−4x−12∣=∣−16∣=16|x^2 - 4x - 12| = |-16| = 16. At the endpoints x=−3x = -3 and x=5x = 5, the values are ∣9∣=9|9| = 9 and ∣−7∣=7|-7| = 7, respectively. Thus, the maximum possible value of ∣x2−4x−12∣|x^2 - 4x - 12| on the interval is 16.

Step-by-Step Solution

1
Determine the solution set of the absolute value equation ∣x+3∣+∣x−5∣=8|x + 3| + |x - 5| = 8.
The domain of valid xx values is the continuous closed interval [−3,5][-3, 5].
By the geometric distance interpretation, the sum of distances from xx to −3-3 and xx to 55 equals the total distance between −3-3 and 55 (which is 88) if and only if xx lies between −3-3 and 55 inclusive.
2
Express the quadratic expression g(x)=x2−4x−12g(x) = x^2 - 4x - 12 in vertex form.
g(x)=(x−2)2−16g(x) = (x - 2)^2 - 16.
Completing the square allows straightforward evaluation of the vertex and minimum/maximum values of the quadratic on [−3,5][-3, 5].
3
Find the range of g(x)=(x−2)2−16g(x) = (x - 2)^2 - 16 for x∈[−3,5]x \in [-3, 5].
The minimum value occurs at the vertex x=2x = 2, where g(2)=−16g(2) = -16. The maximum value occurs at the endpoint x=−3x = -3, where g(−3)=9g(-3) = 9. Thus, −16≤g(x)≤9-16 \le g(x) \le 9.
The parabola opens upwards with vertex at x=2∈[−3,5]x = 2 \in [-3, 5]. The distance from x=2x = 2 to x=−3x = -3 is 55, while the distance to x=5x = 5 is 33.
4
Evaluate the maximum value of ∣g(x)∣=∣x2−4x−12∣|g(x)| = |x^2 - 4x - 12| on [−3,5][-3, 5].
The absolute value ∣g(x)∣|g(x)| ranges from 00 to max⁡(∣−16∣,∣9∣)=16\max(|-16|, |9|) = 16.
The absolute value converts negative outputs to positive, so the extreme magnitude ∣−16∣=16|-16| = 16 at x=2x = 2 is the maximum value.

Key Concept

Absolute Value Distance Interpretation and Quadratic Range Optimization
Estimated Time:2m 0s
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Topics

Inequalities and Absolute Value Equations

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