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

Last updated at March 11, 2021 by Teachoo

Last updated at March 11, 2021 by Teachoo

Transcript

Ex 5.8, 1 Verify Rolleβs theorem for the function π (π₯) = π₯2 + 2π₯ β 8, π₯ β [β 4, 2]. π (π₯) = π₯2 + 2π₯ β 8, π₯ β [β 4, 2]. Rolleβs theorem is satisfied if Condition 1 π(π₯)=π₯2 + 2π₯ β 8 is continuous at (β4 , 2) Since π(π₯)=π₯2 + 2π₯ β 8 is a polynomial & Every polynomial function is continuous for all π₯ βπ β π(π₯)=π₯2 + 2π₯ β 8 is continuous at π₯β[β 4, 2] Conditions of Rolleβs theorem π(π₯) is continuous at (π , π) π(π₯) is derivable at (π , π) π(π) = π(π) If all 3 conditions satisfied then there exist some c in (π , π) such that πβ²(π) = 0 Condition 2 π(π₯)=π₯2 + 2π₯ β 8 is differentiable at (β4 , 2) π(π₯) =π₯2 + 2π₯ β 8 is a polynomial . & Every polynomial function is differentiable for all π₯ βπ therefore π(π₯) is differentiable at (β4 , 2) Condition 3 π(π₯) = π₯2 + 2π₯ β 8" " π(β4) = (β4)^2+2(β4)β8 = 16 β 8 β 8 = 16 β 16 = 0 & π(2) = (2)^2+2(2)β8 = 4+4β8 = 8β8 = 0 Conditions of Rolleβs theorem π(π₯) is continuous at (π , π) π(π₯) is derivable at (π , π) π(π) = π(π) If all 3 conditions satisfied then there exist some c in (π , π) such that πβ²(π) = 0 Hence π(β4) = π(2) Now, π(π₯) = π₯2 + 2π₯ β 8" " π^β² (π₯) = 2π₯+2β0 π^β² (π₯) = 2π₯+2 π^β² (π₯) = 2π+2 Since all three condition satisfied π^β² (π) = 0 2π+2 = 0 2c = β 2 Conditions of Rolleβs theorem π(π₯) is continuous at (π , π) π(π₯) is derivable at (π , π) π(π) = π(π) If all 3 conditions satisfied then there exist some c in (π , π) such that πβ²(π) = 0 π = (β 2)/2 = β1 Value of c = β1 β(β4 , 2) Thus, Rolleβs Theorem is satisfied.

Ex 5.8

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.