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Example 42 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)
Last updated at April 13, 2021 by Teachoo
Verify Rolles theorem
Chapter 5 Class 12 Continuity and Differentiability
Concept wise
- Checking continuity at a given point
- Checking continuity at any point
- Checking continuity using LHL and RHL
- Algebra of continous functions
- Continuity of composite functions
- Checking if funciton is differentiable
- Finding derivative of a function by chain rule
- Finding derivative of Implicit functions
- Finding derivative of Inverse trigonometric functions
- Finding derivative of Exponential & logarithm functions
- Logarithmic Differentiation - Type 1
- Logarithmic Differentiation - Type 2
- Derivatives in parametric form
- Finding second order derivatives - Normal form
- Finding second order derivatives- Implicit form
- Proofs
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Verify Rolles theorem
- Verify Mean Value Theorem
Transcript
Example 42 Verify Rolleβs theorem for the function y = x2 + 2, a = β 2 and b = 2. y = x2 + 2, a = β2 and b = 2 Let π(π₯) = π₯^2+2 Rolleβs theorem is satisfied if Condition 1 Since π(π₯) is a polynomial, it is continuous β΄ π(π₯) is continuous at (β2 , 2) Conditions of Rolleβs theorem π(π₯) is continuous at (π , π) π(π₯) is differentiable 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 Since π(π₯) is a polynomial, it is differentiable β΄ π(π₯) is differentiable at (β2 , 2) Also, π(βπ) = (β2)^2+2 = 4+2 = 6 π(π) = 2^2+2 = 4+2 = 6 Hence, π(π) = π(βπ) Conditions of Rolleβs theorem π(π₯) is continuous at (π , π) π(π₯) is differentiable 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 Also, π(π₯) = π₯^2+2 π^β² (π₯) = 2x So, π^β² (π) = ππ Since all 3 conditions are satisfied π^β² (π) = π 2π = 0 π = π Value of c i.e. 0 lies between β2 and 2. Hence c = 0 β (βπ, π) Thus, Rolleβs theorem is satisfied.
