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Last updated at April 19, 2021 by Teachoo

Example 33 (Method 1) Find all the points of local maxima and local minima of the function f given by π(π₯)=2π₯3 β6π₯2+6π₯+5. π(π₯)=2π₯3 β6π₯2+6π₯+5 Finding fβ(π) πβ(π₯)=π(2π₯3 β 6π₯2 + 6π₯ + 5)/ππ₯ πβ(π₯)=6π₯^2β12π₯+6+0 πβ(π₯)=6(π₯^2β2π₯+1) πβ(π₯)=6((π₯)^2+(1)^2β2(π₯)(1)) πβ(π₯)=π(πβπ)^π Putting fβ(π)=π 6(π₯β1)^2=0 (π₯β1)^2=0 π₯β1=0 So, π=π is only critical point Hence π=π is point of inflexion Example 33 (Method 2) Find all the points of local maxima and local minima of the function f given by π(π₯)=2π₯3 β6π₯2+6π₯+5. π(π₯)=2π₯3 β6π₯2+6π₯+5 Finding fβ(π) πβ(π₯)=π(2π₯3 β 6π₯2 + 6π₯ + 5" " )/ππ₯ πβ(π₯)=6π₯^2β12π₯+6++0 πβ(π₯)=6(π₯^2β2π₯+1) πβ(π₯)=6((π₯)^2+(1)^2β2(π₯)(1)) πβ(π₯)=π(πβπ)^π Putting fβ(π)=π 6(π₯β1)^2=0 (π₯β1)^2=0 So, π=π is the only critical point Finding fββ(π) fββ(π₯)=6.(π(π₯ β 1)^2)/ππ₯ fββ(π₯)=6 Γ 2(π₯β1) fββ(π₯) = 12 (π₯β1) Putting π=π fββ(π₯)=12(1β1) = 0 Since fββ(1) = 0 Hence, π₯=1 is neither point of Maxima nor point of Minima β΄ π=π is Point of Inflexion.