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Let f(x) =|x-2| +|2.5-x|+|3.6-x|, where ...

Let `f(x) =|x-2| +|2.5-x|+|3.6-x|`, where x is a real number, attains a minimum at?

A

x=2.3

B

x=2.5

C

x=2.7

D

None of these

Text Solution

AI Generated Solution

The correct Answer is:
To find the minimum of the function \( f(x) = |x - 2| + |2.5 - x| + |3.6 - x| \), we can follow these steps: ### Step 1: Identify the critical points The critical points for the function occur where the expressions inside the absolute values are equal to zero. Therefore, we set each expression to zero: 1. \( x - 2 = 0 \) → \( x = 2 \) 2. \( 2.5 - x = 0 \) → \( x = 2.5 \) 3. \( 3.6 - x = 0 \) → \( x = 3.6 \) So, the critical points are \( x = 2, 2.5, 3.6 \). ### Step 2: Evaluate the function at the critical points We will evaluate \( f(x) \) at each of these critical points: 1. For \( x = 2 \): \[ f(2) = |2 - 2| + |2.5 - 2| + |3.6 - 2| = 0 + 0.5 + 1.6 = 2.1 \] 2. For \( x = 2.5 \): \[ f(2.5) = |2.5 - 2| + |2.5 - 2.5| + |3.6 - 2.5| = 0.5 + 0 + 1.1 = 1.6 \] 3. For \( x = 3.6 \): \[ f(3.6) = |3.6 - 2| + |2.5 - 3.6| + |3.6 - 3.6| = 1.6 + 1.1 + 0 = 2.7 \] ### Step 3: Compare the values Now we compare the values of \( f(x) \) at the critical points: - \( f(2) = 2.1 \) - \( f(2.5) = 1.6 \) - \( f(3.6) = 2.7 \) The minimum value occurs at \( x = 2.5 \) where \( f(2.5) = 1.6 \). ### Conclusion Thus, the function \( f(x) \) attains its minimum at \( x = 2.5 \). ---
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