Question:
Consider the real-valued function \( g(x)=\max\{(x-2)^2,\,-2x+7\}\), \(x\in(-\infty,\infty)\). The minimum value attained by \(g(x)\) is _____(rounded off to one decimal place).
Consider the real-valued function \( g(x)=\max\{(x-2)^2,\,-2x+7\}\), \(x\in(-\infty,\infty)\). The minimum value attained by \(g(x)\) is _____(rounded off to one decimal place).
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For \(g(x)=\max\{f_1(x),f_2(x)\}\), the minimum typically occurs where \(f_1=f_2\) (the βequalizerβ point).
Updated On: Sep 1, 2025
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Correct Answer: 0.9
Solution and Explanation
Step 1: Find where the two branches are equal
Solve \((x-2)^2=-2x+7\Rightarrow x^2-2x-3=0\Rightarrow x=\{-1,\,3\}\).
Step 2: Evaluate \(g(x)\) at the intersections
At \(x=-1\): both branches give \(9\). At \(x=3\): both give \(1\).
Step 3: Conclude the minimum of the max
Between intersections the larger curve switches; the least possible common value is at \(x=3\). Hence \(\min g(x)=1\).
Solve \((x-2)^2=-2x+7\Rightarrow x^2-2x-3=0\Rightarrow x=\{-1,\,3\}\).
Step 2: Evaluate \(g(x)\) at the intersections
At \(x=-1\): both branches give \(9\). At \(x=3\): both give \(1\).
Step 3: Conclude the minimum of the max
Between intersections the larger curve switches; the least possible common value is at \(x=3\). Hence \(\min g(x)=1\).
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