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

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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﷯ = −4﷯﷮2﷯+2 −4﷯−8 = 16 − 8 − 8 = 16 − 16 = 0 & 𝑓(2) = 2﷯﷮2﷯+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 = − 2﷮2﷯ = −1 Value of c = −1 ∈ −4 , 2﷯ Thus, Rolle’s Theorem is satisfied.