Verify Rolles theorem

Chapter 5 Class 12 Continuity and Differentiability
Concept wise

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Ex 5.8, 1 Verify Rolleβs theorem for the function π (π₯) = π₯2 + 2π₯ β 8, π₯ β [4, 2].Letβs check conditions of Rolleβs theorem Condition 1 We need to check if π(π₯) is continuous at [β4, 2] Since π(π)=π₯2 + 2π₯ β 8 is a polynomial & Every polynomial function is continuous for all π₯ βπ β΄ π(π₯)is continuous at π₯β[β4, 2] Conditions of Rolleβs theorem π(π₯) is continuous at [π , π] π(π₯) is derivable at (π , π) π(π) = π(π) If all 3 conditions are satisfied then there exist some c in (π , π) such that πβ²(π) = 0 Condition 2 We need to check if π(π₯) is differentiable at (β4 , 2) Since π(π) =π₯2 + 2π₯ β 8 is a polynomial . & Every polynomial function is differentiable for all π₯ βπ β΄ π(π₯) is differentiable at (β4 , 2) Condition 3 We need to check if π(π) = π(π), for a = β4, b = 2 π(βπ) π(βπ) = (β4)^2+2(β4)β8 = 16 β 8 β 8 = 0 π(π) π(π)" = " (2)^2+2(2)β8" " "= " 4+4β8" = 0" Hence, π(β4) = π(2) Now, π(π₯) = π₯2 + 2π₯ β 8" " π^β² (π) = 2π₯+2β0 π^β² (π₯) = 2π₯+2 π^β² (π) = ππ+π Since all three conditions satisfied π^β² (π) = π 2π+2 = 0 2c = β 2 c = (β2)/2 c = β1 Since c = β1 β(β4 , 2) Thus, Rolleβs Theorem is satisfied.