    1. Chapter 5 Class 12 Continuity and Differentiability (Term 1)
2. Serial order wise
3. Ex 5.8

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] Conditions of Rolle’s theorem 𝑓(𝑥) is continuous at (𝑎 , 𝑏) 𝑓(𝑥) is derivable at (𝑎 , 𝑏) 𝑓(𝑎) = 𝑓(𝑏) If all 3 conditions satisfied then there exist some c in (𝑎 , 𝑏) such that 𝑓′(𝑐) = 0 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) = (−4)^2+2(−4)−8 = 16 − 8 − 8 = 16 − 16 = 0 & 𝑓(2) = (2)^2+2(2)−8 = 4+4−8 = 8−8 = 0 Conditions of Rolle’s theorem 𝑓(𝑥) is continuous at (𝑎 , 𝑏) 𝑓(𝑥) is derivable at (𝑎 , 𝑏) 𝑓(𝑎) = 𝑓(𝑏) If all 3 conditions satisfied then there exist some c in (𝑎 , 𝑏) such that 𝑓′(𝑐) = 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 Conditions of Rolle’s theorem 𝑓(𝑥) is continuous at (𝑎 , 𝑏) 𝑓(𝑥) is derivable at (𝑎 , 𝑏) 𝑓(𝑎) = 𝑓(𝑏) If all 3 conditions satisfied then there exist some c in (𝑎 , 𝑏) such that 𝑓′(𝑐) = 0 𝑐 = (− 2)/2 = −1 Value of c = −1 ∈(−4 , 2) Thus, Rolle’s Theorem is satisfied.

Ex 5.8

Chapter 5 Class 12 Continuity and Differentiability (Term 1)
Serial order wise 