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Example 30 - Chapter 6 Class 12 Application of Derivatives (Term 1)
Last updated at May 6, 2021 by Teachoo
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Chapter 6 Class 12 Application of Derivatives
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Transcript
Example 30 (Method 1) Find all the points of local maxima and local minima of the function f given by f (π₯)=2π₯3 β6π₯2+6π₯+5.f (π₯)=2π₯3 β6π₯2+ 6π₯+5 Finding fβ (π) f β²(π₯)= π(2π₯3 β 6π₯2 + 6π₯ + 5)/ππ₯ f β²(π₯)=6π₯2 β12π₯+6+0 f β²(π₯)=6(π₯^2β2π₯+1) Putting f β²(π)= 0 6(π₯^2β2π₯+1)=0 π₯^2β2π₯+1=0 (π₯)^2+(1)^2β2(π₯)(1)=0 (π₯β1)^2=0 So, π=π is only critical point Hence π=π is point of inflexion Example 30 (Method 2) Find all the points of local maxima and local minima of the function f given by f (π₯)=2π₯3 β6π₯2+6π₯+5. f (π₯)=2π₯3 β6π₯2+ 6π₯+5 Finding fβ (π) f β²(π₯)= π(2π₯3 β 6π₯2+ 6π₯ + 5)/ππ₯ f β²(π₯)=6π₯2 β12π₯+6+0 f β²(π₯)=6(π₯^2β2π₯+1) Putting f β²(π)= 0 6(π₯^2β2π₯+1)=0 π₯^2β2π₯+1=0 π₯^2+1^2β2(π₯)(1)=0 (π₯β1)^2=0 So, π=π is only critical point Finding fββ(π) fββ(π₯)=6 π(π₯^2 β 2π₯ + 1)/ππ₯ fββ(π₯)=6(2π₯β2+0) fββ(π₯)=12(π₯β1) Putting π=π fββ(1)=12(1β1) = 12 Γ 0 = 0 Since fββ(1) = 0 Hence, π₯=1 is neither point of Maxima nor point of Minima β΄ π=π is Point of Inflexion.
