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Last updated at March 10, 2021 by Teachoo

Ex 5.1, 7 Find all points of discontinuity of f, where f is defined by π(π₯)={β(|π₯|+3, ππ π₯β€β3@ β2π₯, ππβ3<π₯<3@ 6π₯+2, ππ π₯β₯3)β€ Since we need to find continuity at of the function We check continuity for different values of x When x < β3 When x = β3 When β3 < x < 3 When x = 3 When x > 3 Case 1 : When x < β3 For x < β3, f(x) = |π₯|+3 f(x) = βx + 3 Since this a polynomial It is continuous β΄ f(x) is continuous for x < β3 (As x < β3, x is negative) Case 2 : When x = β3 f(x) is continuous at π₯ =β3 if L.H.L = R.H.L = π(β3) if limβ¬(xβγβ3γ^β ) π(π₯)=limβ¬(xβγβ3γ^+ ) " " π(π₯)= π(β3) Since there are two different functions on the left & right of β3, we take LHL & RHL . LHL at x β β3 limβ¬(xβγβ3γ^β ) f(x) = limβ¬(hβ0) f(β3 β h) = limβ¬(hβ0) (|β3ββ|+3) = |β3β0|+3 = |β3|+3 = 3 + 3 = 6 RHL at x β β3 limβ¬(xβγβ3γ^+ ) f(x) = limβ¬(hβ0) f(β3 + h) = limβ¬(hβ0) β2(β3+β) = limβ¬(hβ0) 6β2β = 6 β 0 = 6 & π(βπ) = |β3|+3 = 3+3 = 6 Hence, L.H.L = R.H.L = π(β3) β΄ f is continuous at x = β3 Case 3 : When β3 < x < 3 For β3 < x < 3, f(x) = β2x Since this a polynomial It is continuous β΄ f(x) is continuous for β3 < x < 3 Case 4 : When x = 3 f(x) is continuous at π₯ =3 if L.H.L = R.H.L = π(3) if limβ¬(xβ3^β ) π(π₯)=limβ¬(xβ3^+ ) " " π(π₯)= π(3) Since there are two different functions on the left & right of 3, we take LHL & RHL . LHL at x β 3 limβ¬(xβ3^β ) f(x) = limβ¬(hβ0) f(3 β h) = limβ¬(hβ0) β2(3βh) = limβ¬(hβ0) β6+2h = β6 + 0 = β6 RHL at x β 3 limβ¬(xβ3^+ ) f(x) = limβ¬(hβ0) f(3 + h) = limβ¬(hβ0) 6(3+β)+2 = limβ¬(hβ0) 18+6β+2 = limβ¬(hβ0) 20+6β = 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 β΄ 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 βπβ{π}