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प्रश्न
Redefine the function f(x) = x − 2 + 2 + x , – 3 ≤ x ≤ 3
योग
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उत्तर
We know that
when x > 0
|x – 2| is (x – 2), x ≥ 2
|2 + x| is (2 + x), x ≥ –2
when x > 0
|x – 2| is –(x – 2), x < 2
|2 + x| is –(2 + x), x <–2
Given that, f(x) = |x – 2| + |2 + x|, –3 ≤ x ≤ 3
It can be rewritten as,
f(x) = `{{:(-(x - 2) - (2 + x)",", -3 ≤ x < - 2),(-(x - 2) + (2 + x)",", -2 ≤ x < 2),((x - 2) + (2 + x)",", 2 ≤ x ≤ 3):}`
Or
f(x) = `{{:(-x + 2 - 2 - x",", -3 ≤ x < -2),(-x + 2 + 2 + x",", -2 ≤ x < 2),(x - 2 + 2 + x",", 2 ≤ x ≤ 3):}`
Or
f(x) = `{{:(-2x",", -3 ≤ x < -2),(4",", -2 ≤ x < 2),(2x",", 2 ≤ x ≤ 3):}`
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