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What is the range of (f(x)=2|x|+3)?

Answer and explanation

Correct answer: ([3,\infty))

The direct answer is A: the range is \([3,\infty)\). Start with the basic fact \(|x|\ge0\) for every real x. Multiplying by 2 gives \(2|x|\ge0\), and adding 3 gives \(2|x|+3\ge3\). Thus no output is below 3. The minimum 3 is reached at x=0, because f(0)=2|0|+3=3. As |x| can take every non-negative value, 2|x| can take every non-negative value, so all values from 3 upward are attained. Option A is correct. Option B, [0,∞), has an incorrect lower endpoint; outputs between 0 and 3 cannot occur. Option C, (−∞,3], gives the wrong direction and includes impossible negative values. Option D, all real numbers, is wrong for the same reason: the output is never less than 3. The square brackets matter because 3 is actually obtained. Memory cue: absolute value is never negative; positive scaling keeps that fact, and +3 moves the minimum to 3.

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Real-Valued-FunctionsRangeModulus-Function

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