Continuity of composite functions

Chapter 5 Class 12 Continuity and Differentiability
Concept wise

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Example 20 Show that the function f defined by f (x) = |1β π₯ + | π₯ ||, where x is any real number is a continuousGiven π(π₯) = |(1βπ₯+|π₯|)| Let π(π) = 1βπ₯+|π₯| & π(π) = |π₯| Then , πππ(π) = β(π(π₯)) = β(1βπ₯+|π₯|) = |(1βπ₯+|π₯|)| = π(π) We know that, Modulus function is continuous β΄ π(π) = |π₯| is continuous Also, π(π) = (πβπ)+|π| Since (1βπ₯) is a polynomial & every polynomial function is continuous β΄ (πβπ) is continuous Also, |π| is also continuous Since Sum of two continuous function is also continuous Thus, π(π₯) = 1βπ₯+|π₯| is continuous . Hence, π(π₯) & β(π₯) are both continuous . We know that If two function of π(π₯) & β(π₯) both continuous, then their composition πππ(π) is also continuous Hence, π(π) is continuous .