Check sibling questions

Ex 5.1, 34 - Find all points of discontinuity f(x) = |x| - |x+1|

Ex 5.1, 34 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.1, 34 - Chapter 5 Class 12 Continuity and Differentiability - Part 3
Ex 5.1, 34 - Chapter 5 Class 12 Continuity and Differentiability - Part 4
Ex 5.1, 34 - Chapter 5 Class 12 Continuity and Differentiability - Part 5
Ex 5.1, 34 - Chapter 5 Class 12 Continuity and Differentiability - Part 6
Ex 5.1, 34 - Chapter 5 Class 12 Continuity and Differentiability - Part 7
Ex 5.1, 34 - Chapter 5 Class 12 Continuity and Differentiability - Part 8
Ex 5.1, 34 - Chapter 5 Class 12 Continuity and Differentiability - Part 9

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Ex 5.1, 34 Find all the points of discontinuity of f defined by 𝑓(π‘₯)= |π‘₯| – |π‘₯+1|.Given 𝑓(π‘₯)= |π‘₯| – |π‘₯+1|. Here, we have 2 critical points x = 0 and x + 1 = 0 i.e. x = 0, and x = βˆ’1 So, our intervals will be When π’™β‰€βˆ’πŸ When βˆ’πŸ<𝒙<𝟎 When 𝒙β‰₯𝟎 When π’™β‰€βˆ’πŸ 𝑓(π‘₯)= |π‘₯| – |π‘₯+1|. Here, both will be negative 𝑓(π‘₯)=(βˆ’π‘₯) –(βˆ’(π‘₯+1)) 𝑓(π‘₯)=βˆ’π‘₯+(π‘₯+1) " " 𝒇(𝒙)=𝟏 When βˆ’πŸ<π’™β‰€πŸŽ 𝑓(π‘₯)= |π‘₯| – |π‘₯+1|. Here, x will be negative, but (x + 1) will be positive 𝑓(π‘₯)=(βˆ’π‘₯) –(π‘₯+1) 𝑓(π‘₯)=βˆ’π‘₯βˆ’π‘₯βˆ’1 " " 𝒇(𝒙)=βˆ’πŸπ’™βˆ’πŸ |π‘₯| = {β–ˆ(π‘₯, π‘₯ β‰₯0@βˆ’π‘₯, π‘₯<0)─ |π‘₯+1| = {β–ˆ((π‘₯+1) , π‘₯+1β‰₯0@βˆ’(π‘₯+1) π‘₯+1<0)─ = {β–ˆ((π‘₯+1) , π‘₯β‰₯βˆ’1@βˆ’(π‘₯+1) π‘₯<1)─ When 𝒙β‰₯𝟎 𝑓(π‘₯)= |π‘₯| – |π‘₯+1|. Here, both will be positive 𝑓(π‘₯)=π‘₯ –(π‘₯+1) 𝑓(π‘₯)=π‘₯βˆ’π‘₯βˆ’1 " " 𝒇(𝒙)=βˆ’πŸ Thus, our function becomes 𝒇(𝒙)={β–ˆ(𝟏 π’Šπ’‡ π’™β‰€βˆ’πŸ@βˆ’πŸπ’™βˆ’πŸ π’Šπ’‡ βˆ’πŸ<𝒙<𝟎@βˆ’πŸ π’Šπ’‡ 𝒙β‰₯𝟎)─ 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 < 0 When x = 0 When x > 0 Checking continuity Case 1 : When x < βˆ’1 For x < βˆ’1, f(x) = 1 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) 1 = 1 RHL at x β†’ 0 lim┬(xβ†’γ€–βˆ’1γ€—^+ ) f(x) = lim┬(hβ†’0) f(βˆ’1 + h) = lim┬(hβ†’0) (βˆ’2(βˆ’1+β„Ž))βˆ’1 = lim┬(hβ†’0) (2βˆ’2β„Ž)βˆ’1 = (2 βˆ’ 2(0)) βˆ’ 1 = 2 βˆ’ 0 βˆ’ 1 = 1 & 𝒇(βˆ’πŸ) = 1 Hence, L.H.L = R.H.L = 𝑓(βˆ’1) ∴ f is continuous at x = βˆ’1 Case 3 : When βˆ’1 < x < 0 For βˆ’1 < x < 0 f(x) = βˆ’2x βˆ’ 1 Since this a polynomial It is continuous ∴ f(x) is continuous for βˆ’1 < x < 0 Case 4 : When x = 0 f(x) is continuous at π‘₯ =0 if L.H.L = R.H.L = 𝑓(0) if lim┬(xβ†’0^βˆ’ ) 𝑓(π‘₯)=lim┬(xβ†’0^+ ) " " 𝑓(π‘₯)= 𝑓(0) Since there are two different functions on the left & right of 0, we take LHL & RHL . LHL at x β†’ 0 lim┬(xβ†’0^βˆ’ ) f(x) = lim┬(hβ†’0) f(0 βˆ’ h) = lim┬(hβ†’0) f(βˆ’h) = lim┬(hβ†’0) βˆ’2(βˆ’β„Ž)βˆ’1 = lim┬(hβ†’0) 2β„Žβˆ’1 = 2(0) βˆ’ 1 = βˆ’1 RHL at x β†’ 0 lim┬(xβ†’0^+ ) f(x) = lim┬(hβ†’0) f(0 + h) = lim┬(hβ†’0) f(h) = lim┬(hβ†’0) βˆ’1 = βˆ’1 & 𝑓(0) = βˆ’1 Hence, L.H.L = R.H.L = 𝑓(0) ∴ f is continuous at x = 0 Case 5: When x > 0 For x > 0, f(x) = βˆ’1 Since this constant It is continuous ∴ f(x) is continuous for x > 0 Since there is no point of discontinuity Therefore, f is continuous for all x ∈ R

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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.