Last updated at Dec. 16, 2024 by Teachoo
Question 1 Verify Rolleβs theorem for the function π (π₯) = π₯2 + 2π₯ β 8, π₯ β [4, 2].Letβs check conditions of Rolleβs theorem Condition 1 We need to check if π(π₯) 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₯)is continuous at π₯β[β4, 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 are satisfied then there exist some c in (π , π) such that πβ²(π) = 0 Condition 2 We need to check if π(π₯) is differentiable at (β4 , 2) Since π(π) =π₯2 + 2π₯ β 8 is a polynomial . & Every polynomial function is differentiable for all π₯ βπ β΄ π(π₯) is differentiable at (β4 , 2) Condition 3 We need to check if π(π) = π(π), for a = β4, b = 2 π(βπ) π(βπ) = (β4)^2+2(β4)β8 = 16 β 8 β 8 = 0 π(π) π(π)" = " (2)^2+2(2)β8" " "= " 4+4β8" = 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) = ππ+π Since all three conditions satisfied π^β² (π) = π 2π+2 = 0 2c = β 2 c = (β2)/2 c = β1 Since c = β1 β(β4 , 2) Thus, Rolleβs Theorem is satisfied.
About the Author
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo