Example 20 - Show that f(x) = |1 - x + |x|| is continuous - Examples

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  1. Chapter 5 Class 12 Continuity and Differentiability
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Example 20 Show that the function f defined by f (x) = |1βˆ’ π‘₯ + | π‘₯ ||, where x is any real number is a continuous 𝑓﷐π‘₯ο·― = ﷐1βˆ’π‘₯+﷐π‘₯ο·―ο·― Let 𝑔(π‘₯) = 1βˆ’π‘₯+﷐π‘₯ο·― & β„Ž(π‘₯) = ﷐π‘₯ο·― Then ο·β„Ž π‘œ 𝑔﷯﷐π‘₯ο·― = β„Žο·π‘”ο·π‘₯ο·―ο·― = β„Žο·1βˆ’π‘₯+﷐π‘₯ο·―ο·― = ﷐1βˆ’π‘₯+﷐π‘₯ο·―ο·― Now β„Ž(π‘₯) = ﷐π‘₯ο·― We know that Modulus function is continuous β‡’ β„Ž(π‘₯) = ﷐π‘₯ο·― is continuous 𝑔(π‘₯) = ﷐1βˆ’π‘₯ο·―+﷐π‘₯ο·― Since ﷐1βˆ’π‘₯ο·― is a polynomial & we know that every polynomial function is continuous β‡’ ﷐1βˆ’π‘₯ο·― is continuous & β„Ž(π‘₯) = ﷐π‘₯ο·― is also continuous We know that Sum of two continuous function is also continuous 𝑔﷐π‘₯ο·― = 1βˆ’π‘₯+﷐π‘₯ο·― is continuous . Hence 𝑔﷐π‘₯ο·― & β„Žο·π‘₯ο·― are both continuous . If two function of 𝑔﷐π‘₯ο·― & β„Žο·π‘₯ο·― both continuous then their composition ο·β„Ž π‘œ 𝑔﷯﷐π‘₯ο·― is also continuous Hence 𝒇﷐𝒙﷯ is continuous .

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