The minimum value of the function \(f(x) = |x| + |2x+3|\) for real \(x\) is ________ (rounded off to 1 decimal place).
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Correct Answer: 1.5
Solution and Explanation
The function \(f(x) = |x| + |2x+3|\) changes its algebraic form wherever an expression inside a modulus sign changes sign. That happens at \(x = 0\) and at \(2x+3=0\), i.e. \(x = -3/2\). These two points split the real line into three regions.
Step 2: Write \(f(x)\) in each region.
For \(x \geq 0\): both \(x\) and \(2x+3\) are non-negative, so
\[ f(x) = x + (2x+3) = 3x + 3 \]
This is increasing in \(x\) (slope \(+3\)), so on this region the smallest value is at \(x=0\): \(f(0)=3\).
For \(-3/2 \leq x < 0\): here \(x<0\) but \(2x+3 \geq 0\), so
\[ f(x) = -x + (2x+3) = x + 3 \]
This is also increasing (slope \(+1\)), so the smallest value on this region is at the left end \(x=-3/2\): \(f(-3/2) = -3/2+3 = 1.5\).
For \(x < -3/2\): here both \(x\) and \(2x+3\) are negative, so
\[ f(x) = -x - (2x+3) = -3x - 3 \]
This has slope \(-3\) (decreasing as \(x\) increases), so as \(x\) moves from \(-\infty\) up to \(-3/2\), \(f(x)\) keeps falling and reaches its smallest value on this region right at \(x=-3/2\), which again gives \(f(-3/2)=1.5\).
Step 3: Compare the three regions.
The candidate minimum values are \(f(0)=3\) and \(f(-3/2)=1.5\) (the function is monotonic on every region, so the overall minimum can only occur at a break point). The smaller of these is \(1.5\).
Final Answer:
\[ \boxed{f(x)_{min} = 1.5} \]
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