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Ex 5.1, 16 Discuss the continuity of the function f, where f is defined by 𝑓(π‘₯)={β–ˆ(βˆ’&2, 𝑖𝑓 π‘₯β‰€βˆ’1@2&π‘₯, 𝑖𝑓 βˆ’1<π‘₯≀1@2, 𝑖𝑓 π‘₯>1 )─ Since we need to find continuity at of the function We check continuity for different values of x When x < βˆ’1 When x = βˆ’1 When βˆ’1 < x < 1 When x = 1 When x > 1 Case 1 : When x < βˆ’1 For x < βˆ’1, f(x) = βˆ’2 Since this constant It is continuous ∴ f(x) is continuous for x < βˆ’1 Case 2 : When x = βˆ’1 f(x) is continuous at π‘₯ =βˆ’1 if L.H.L = R.H.L = 𝑓(βˆ’1) if lim┬(xβ†’γ€–βˆ’1γ€—^βˆ’ ) 𝑓(π‘₯)=lim┬(xβ†’γ€–βˆ’1γ€—^+ ) " " 𝑓(π‘₯)= 𝑓(βˆ’1) Since there are two different functions on the left & right of βˆ’1, we take LHL & RHL . LHL at x β†’ βˆ’1 lim┬(xβ†’γ€–βˆ’1γ€—^βˆ’ ) f(x) = lim┬(hβ†’0) f(βˆ’1 βˆ’ h) = lim┬(hβ†’0) βˆ’2 = βˆ’2 RHL at x β†’ 0 lim┬(xβ†’γ€–βˆ’1γ€—^+ ) f(x) = lim┬(hβ†’0) f(βˆ’1 + h) = lim┬(hβ†’0) 2(βˆ’1+β„Ž) = 2(βˆ’1+0) = βˆ’2 & 𝑓(βˆ’1) = βˆ’2 Hence, L.H.L = R.H.L = 𝑓(βˆ’1) ∴ f is continuous at x = βˆ’1 Case 3 : When βˆ’1 < x < 1 For βˆ’1 < x < 1 f(x) = 2x Since this a polynomial It is continuous ∴ f(x) is continuous for βˆ’1 < x < 1 Case 4 : When x = 1 f(x) is continuous at π‘₯ =1 if L.H.L = R.H.L = 𝑓(1) if lim┬(xβ†’1^βˆ’ ) 𝑓(π‘₯)=lim┬(xβ†’1^+ ) " " 𝑓(π‘₯)= 𝑓(1) & 𝑓(1) = 2(1) = 2 Since there are two different functions on the left & right of 1, we take LHL & RHL . LHL at x β†’ 1 lim┬(xβ†’1^βˆ’ ) f(x) = lim┬(hβ†’0) f(1 βˆ’ h) = lim┬(hβ†’0) 2(1βˆ’β„Ž) = 2(1βˆ’0) = 2 RHL at x β†’ 1 lim┬(xβ†’1^+ ) f(x) = lim┬(hβ†’0) f(1 + h) = lim┬(hβ†’0) 2 = 2 Hence, L.H.L = R.H.L = 𝑓(1) ∴ f is continuous at x=1 Case 5: When x > 1 For x > 1, f(x) = 2 Since this constant It is continuous ∴ f(x) is continuous for x > 1 Thus, f(x) is continuous for all real numbers, i.e. f is continuous for all x ∈ R

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.