Check sibling questions

Ex 5.1, 7 - Find all points of discontinuity of f(x) = {|x| + 3

Ex 5.1 ,7 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.1 ,7 - Chapter 5 Class 12 Continuity and Differentiability - Part 3
Ex 5.1 ,7 - Chapter 5 Class 12 Continuity and Differentiability - Part 4
Ex 5.1 ,7 - Chapter 5 Class 12 Continuity and Differentiability - Part 5
Ex 5.1 ,7 - Chapter 5 Class 12 Continuity and Differentiability - Part 6

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Ex 5.1, 7 Find all points of discontinuity of f, where f is defined by 𝑓(π‘₯)={β–ˆ(|π‘₯|+3, 𝑖𝑓 π‘₯β‰€βˆ’3@ βˆ’2π‘₯, π‘–π‘“βˆ’3<π‘₯<3@ 6π‘₯+2, 𝑖𝑓 π‘₯β‰₯3)─ Since we need to find continuity at of the function We check continuity for different values of x When x < βˆ’3 When x = βˆ’3 When βˆ’3 < x < 3 When x = 3 When x > 3 Case 1 : When x < βˆ’3 For x < βˆ’3, f(x) = |π‘₯|+3 f(x) = βˆ’x + 3 Since this a polynomial It is continuous ∴ f(x) is continuous for x < βˆ’3 (As x < βˆ’3, x is negative) Case 2 : When x = βˆ’3 f(x) is continuous at π‘₯ =βˆ’3 if L.H.L = R.H.L = 𝑓(βˆ’3) if lim┬(xβ†’γ€–βˆ’3γ€—^βˆ’ ) 𝑓(π‘₯)=lim┬(xβ†’γ€–βˆ’3γ€—^+ ) " " 𝑓(π‘₯)= 𝑓(βˆ’3) Since there are two different functions on the left & right of βˆ’3, we take LHL & RHL . LHL at x β†’ βˆ’3 lim┬(xβ†’γ€–βˆ’3γ€—^βˆ’ ) f(x) = lim┬(hβ†’0) f(βˆ’3 βˆ’ h) = lim┬(hβ†’0) (|βˆ’3βˆ’β„Ž|+3) = |βˆ’3βˆ’0|+3 = |βˆ’3|+3 = 3 + 3 = 6 RHL at x β†’ βˆ’3 lim┬(xβ†’γ€–βˆ’3γ€—^+ ) f(x) = lim┬(hβ†’0) f(βˆ’3 + h) = lim┬(hβ†’0) βˆ’2(βˆ’3+β„Ž) = lim┬(hβ†’0) 6βˆ’2β„Ž = 6 βˆ’ 0 = 6 & 𝒇(βˆ’πŸ‘) = |βˆ’3|+3 = 3+3 = 6 Hence, L.H.L = R.H.L = 𝑓(βˆ’3) ∴ f is continuous at x = βˆ’3 Case 3 : When βˆ’3 < x < 3 For βˆ’3 < x < 3, f(x) = βˆ’2x Since this a polynomial It is continuous ∴ f(x) is continuous for βˆ’3 < x < 3 Case 4 : When x = 3 f(x) is continuous at π‘₯ =3 if L.H.L = R.H.L = 𝑓(3) if lim┬(xβ†’3^βˆ’ ) 𝑓(π‘₯)=lim┬(xβ†’3^+ ) " " 𝑓(π‘₯)= 𝑓(3) Since there are two different functions on the left & right of 3, we take LHL & RHL . LHL at x β†’ 3 lim┬(xβ†’3^βˆ’ ) f(x) = lim┬(hβ†’0) f(3 βˆ’ h) = lim┬(hβ†’0) βˆ’2(3βˆ’h) = lim┬(hβ†’0) βˆ’6+2h = βˆ’6 + 0 = βˆ’6 RHL at x β†’ 3 lim┬(xβ†’3^+ ) f(x) = lim┬(hβ†’0) f(3 + h) = lim┬(hβ†’0) 6(3+β„Ž)+2 = lim┬(hβ†’0) 18+6β„Ž+2 = lim┬(hβ†’0) 20+6β„Ž = 20 + 0 = 20 Since L.H.L β‰  R.H.L f(x) is not continuous at x=3 Case 5: When x > 3 For x > 3, f(x) = 6x + 2 Since this a polynomial It is continuous ∴ f(x) is continuous for x > 3 Hence, f is discontinuous at only π‘₯ =3 Thus, f is continuous at all real numbers except 3. f is continuous at x βˆˆπ‘βˆ’{πŸ‘}

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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.