# Ex 5.8, 1 - Chapter 5 Class 12 Continuity and Differentiability

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 5.8, 1 Verify Rolle’s theorem for the function 𝑓 (𝑥) = 𝑥2 + 2𝑥 – 8, 𝑥 ∈ [– 4, 2]. 𝑓 (𝑥) = 𝑥2 + 2𝑥 – 8, 𝑥 ∈ [– 4, 2]. Rolle’s theorem is satisfied if Condition 1 𝑓 𝑥=𝑥2 + 2𝑥 – 8 is continuous at −4 , 2 Since 𝑓 𝑥=𝑥2 + 2𝑥 – 8 is a polynomial & Every polynomial function is continuous for all 𝑥 ∈𝑅 ⇒ 𝑓 𝑥=𝑥2 + 2𝑥 – 8 is continuous at 𝑥∈[– 4, 2] Condition 2 𝑓 𝑥=𝑥2 + 2𝑥 – 8 is differentiable at −4 , 2 𝑓(𝑥) =𝑥2 + 2𝑥 – 8 is a polynomial . & Every polynomial function is differentiable for all 𝑥 ∈𝑅 therefore 𝑓(𝑥) is differentiable at −4 , 2 Condition 3 𝑓 𝑥 = 𝑥2 + 2𝑥 – 8 𝑓 −4 = −42+2 −4−8 = 16 − 8 − 8 = 16 − 16 = 0 & 𝑓(2) = 22+2 2−8 = 4+4−8 = 8−8 = 0 Hence 𝑓 −4 = 𝑓 2 Now, 𝑓 𝑥 = 𝑥2 + 2𝑥 – 8 𝑓′ 𝑥 = 2𝑥+2−0 𝑓′ 𝑥 = 2𝑥+2 𝑓′ 𝑥 = 2𝑐+2 Since all three condition satisfied 𝑓′ 𝑐 = 0 2𝑐+2 = 0 2c = – 2 𝑐 = − 22 = −1 Value of c = −1 ∈ −4 , 2 Thus, Rolle’s Theorem is satisfied.

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.