# Ex 5.8, 3 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)

Last updated at March 11, 2021 by

Last updated at March 11, 2021 by

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Ex 5.8, 3 If π : [β 5, 5] β π is a differentiable function and if π β²(π₯) does not vanish anywhere, then prove that π (β5) β π (5). π : [β 5, 5] β π is a differentiable β We know that every differentiable function is continuous. Therefore f is continuous & differentiable both on (β5, 5) By Mean Value Theorem There exist some c in (5, β5) Such that π^β² (π)=(π(π) β π(π))/(π β π) Given that π^β² (π₯) does not vanish any where β π^β² (π₯) β 0 for any value of x Thus, π^β² (π) β 0 (π(5) β π(β5))/(5 β(β5) ) β 0 (π(5) β π(β5))/(5 + 5) β 0 π(5)β π(β5) β 0 Γ 10 π(5)β π(β5) β 0 π(5) "β " π(β5) Hence proved.

Ex 5.8

Ex 5.8, 1
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