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Difficulty: HardInequalities and Absolute Value Equations

If xx is a real number satisfying the inequality ∣x−3∣+2∣x+1∣≤7|x - 3| + 2|x + 1| \leq 7, what is the minimum possible value of x2−2xx^2 - 2x?

  1. −1-1Answer
  2. B
    00
  3. C
    −2-2
  4. D
    33
  5. E
    88

Answer

The minimum possible value of x2−2xx^2 - 2x is −1-1.
The absolute value inequality ∣x−3∣+2∣x+1∣≤7|x - 3| + 2|x + 1| \leq 7 restricts xx to the closed interval [−2,2][-2, 2]. The quadratic function x2−2xx^2 - 2x can be rewritten as (x−1)2−1(x - 1)^2 - 1, which has a minimum value of −1-1 at x=1x = 1. Because x=1x = 1 is inside the interval [−2,2][-2, 2], the minimum possible value is −1-1.

Step-by-Step Solution

1
Identify the critical points of the absolute value expressions to set up case analysis.
The critical points are x=−1x = -1 and x=3x = 3, dividing the real line into three intervals: x<−1x < -1, −1≤x≤3-1 \leq x \leq 3, and x>3x > 3.
Absolute value expressions ∣x−3∣|x - 3| and ∣x+1∣|x + 1| change sign at x=3x = 3 and x=−1x = -1, respectively.
2
Solve the inequality ∣x−3∣+2∣x+1∣≤7|x - 3| + 2|x + 1| \leq 7 across all three cases.
Case 1 (x<−1x < -1): 3−x−2(x+1)≤7  ⟹  1−3x≤7  ⟹  x≥−23 - x - 2(x + 1) \leq 7 \implies 1 - 3x \leq 7 \implies x \geq -2. Combining gives −2≤x<−1-2 \leq x < -1.
Case 2 (−1≤x≤3-1 \leq x \leq 3): 3−x+2(x+1)≤7  ⟹  x+5≤7  ⟹  x≤23 - x + 2(x + 1) \leq 7 \implies x + 5 \leq 7 \implies x \leq 2. Combining gives −1≤x≤2-1 \leq x \leq 2.
Case 3 (x>3x > 3): x−3+2(x+1)≤7  ⟹  3x−1≤7  ⟹  x≤83x - 3 + 2(x + 1) \leq 7 \implies 3x - 1 \leq 7 \implies x \leq \frac{8}{3}. No solutions since 83<3\frac{8}{3} < 3.
The combined solution set is −2≤x≤2-2 \leq x \leq 2.
Piecewise analysis determines the exact range of valid values for xx.
3
Minimize the quadratic expression f(x)=x2−2xf(x) = x^2 - 2x over the interval [−2,2][-2, 2].
Completing the square gives f(x)=(x−1)2−1f(x) = (x - 1)^2 - 1. The vertex occurs at x=1x = 1. Since x=1∈[−2,2]x = 1 \in [-2, 2], the minimum value is f(1)=−1f(1) = -1.
A parabola opening upward achieves its global minimum at its vertex if the vertex lies within the allowed domain.

Key Concept

Solving piecewise absolute value inequalities and finding the extreme values of a quadratic function over a bounded interval.
Estimated Time:2m 0s
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Topics

Inequalities and Absolute Value Equations

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