At which value of x does the function $f(x) = |12 – x| + |x – 3| – |5 – x|$ reach its minimum?

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Official Explanation

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To find the minimum value of the function

f(x) = |12 – x| + |x – 3| – |5 – x|

, we analyze the critical points where the expressions inside the absolute values change. The critical points are at x = 12, x = 3, and x = 5. Evaluating the function at these points:

- For x = 3:

f(3) = |12 - 3| + |3 - 3| - |5 - 3| = 9 + 0 - 2 = 7

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Comments

JoãoJan 10, 2025

Amazing

DEBRUPNov 6, 2024

Great

GarimaJan 17, 2024

thanks

What this question tests

What it tests

Your ability to solve absolute-value equations and inequalities by considering both cases.

Common trap

Dropping the absolute value bars entirely or missing that |x| = k has two solutions when k > 0.

At which value of x does the function f(x)=∣12–x∣+∣x–3∣–∣5–x∣f(x) = |12 – x| + |x – 3| – |5 – x|f(x)=∣12–x∣+∣x–3∣–∣5–x∣ reach its minimum?