# Ex 5.1 ,3

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 5.1, 3 Examine the following functions for continuity. (a) f (x) = x – 5 f (x) = x – 5 Since x – 5 is a polynomial. f is defined for every real number c. Given function is continuous at x = c , where c is a real number if, limx→𝑐 𝑓(𝑥) = 𝑓(𝑐) Since, limx→𝑐 = f (c) So, f is continuous for x = c, where c is a real number ∴ f is continuous for all real numbers ⇒ f is continuous at each x ∈ R Ex 5.1, 3 Examine the following functions for continuity. (b) f (x) = 1𝑥 − 5 , x ≠ 5 If x = 5 f (x) = 15 − 5 = 10 = Hence at x = 5, f (x) is not defined. Given function is continuous at x = c , where c is a real number except 5 if limx→𝑐 𝑓𝑥 = 𝑓𝑐 Hence for all Real number except 5, limx→𝑐 𝑓𝑥 = 𝑓𝑐 f is continuous for all real numbers except 5 ∴ f is continuous at each 𝐱 ∈ R − 𝟓 Ex 5.1, 3 Examine the following functions for continuity. (c) f (x) = 𝑥2 − 25 𝑥 + 5, x ≠ –5 f (x) = 𝑥2 − 25 𝑥 + 5 x ≠ –5 Putting x = –5 f (−5) = (−5)2 − 25 −5 + 5 = 25− 25 −5 + 5 = 00 = undefined Hence At 𝑥 =−5 , 𝑓𝑥 is not defined. Given function is continuous at x = c , where c is a real number except −5 if limx→𝑐 𝑓(𝑥) = 𝑓(𝑐) Since, limx→𝑐 𝑓𝑥 = f (c) f is continuous for all real numbers except −5 ∴ f is continuous at each 𝐱 ∈ R − −𝟓 Ex 5.1, 3 Examine the following functions for continuity. (d) f (x) = |x – 5| f (x) = 𝑥−5 Case 1 Checking continuity At 𝑥 = 5 f is continuous at 𝑥 = 5 if L.H.L = R.H.L = 𝑓5 if limx→5− 𝑓𝑥=limx→5+ 𝑓(𝑥)= 𝑓(5) & 𝑓(5) = 𝑥−5 = 5−5 = 0 = 0 Hence, L.H.L = R.H.L = 𝑓5 ∴ f is continuous at x=5 Case 2 Checking continuity At 𝑥 =𝑐 𝑥 =𝑐 where c > 5 𝑓𝑥= 𝑥−5 𝑓𝑥= x − 5 f is continuous at x=c (c>5) if if limx→𝑐 𝑓𝑥=𝑓(𝑐) Since limx→𝑐 𝑓𝑥=𝑓(𝑐) Hence f is continuous at x=c , c>5 Case 3 Checking continuity At 𝑥 =𝑐 (c < 5) Consider x = c, c < 5 𝑓𝑥 = 𝑥−5 𝑓𝑥 = −𝑥−5 f is continuous at x=c , c<5 if limx→𝑐 𝑓𝑥=𝑓(𝑐) ∴ limx→𝑐 𝑓𝑥=𝑓(𝑐) when c < 5 Hence f is continuous at x=c , (c<5) Hence 𝑓𝑥 = 𝑥−5 is continuous at all points. i.e. f is continuous at 𝒙 ∈ R.

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.