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Example 30 - Find all points of local maxima, minima - CBSE

Example 30 - Chapter 6 Class 12 Application of Derivatives - Part 2

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Example 30 - Chapter 6 Class 12 Application of Derivatives - Part 3

Example 30 - Chapter 6 Class 12 Application of Derivatives - Part 4
Example 30 - Chapter 6 Class 12 Application of Derivatives - Part 5

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Example 30 (Method 1) Find all the points of local maxima and local minima of the function f given by f (π‘₯)=2π‘₯3 –6π‘₯2+6π‘₯+5.f (π‘₯)=2π‘₯3 –6π‘₯2+ 6π‘₯+5 Finding f’ (𝒙) f β€²(π‘₯)= 𝑑(2π‘₯3 – 6π‘₯2 + 6π‘₯ + 5)/𝑑π‘₯ f β€²(π‘₯)=6π‘₯2 –12π‘₯+6+0 f β€²(π‘₯)=6(π‘₯^2βˆ’2π‘₯+1) Putting f β€²(𝒙)= 0 6(π‘₯^2βˆ’2π‘₯+1)=0 π‘₯^2βˆ’2π‘₯+1=0 (π‘₯)^2+(1)^2βˆ’2(π‘₯)(1)=0 (π‘₯βˆ’1)^2=0 So, 𝒙=𝟏 is only critical point Hence 𝒙=𝟏 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 –6π‘₯2+6π‘₯+5. f (π‘₯)=2π‘₯3 –6π‘₯2+ 6π‘₯+5 Finding f’ (𝒙) f β€²(π‘₯)= 𝑑(2π‘₯3 – 6π‘₯2+ 6π‘₯ + 5)/𝑑π‘₯ f β€²(π‘₯)=6π‘₯2 –12π‘₯+6+0 f β€²(π‘₯)=6(π‘₯^2βˆ’2π‘₯+1) Putting f β€²(𝒙)= 0 6(π‘₯^2βˆ’2π‘₯+1)=0 π‘₯^2βˆ’2π‘₯+1=0 π‘₯^2+1^2βˆ’2(π‘₯)(1)=0 (π‘₯βˆ’1)^2=0 So, 𝒙=𝟏 is only critical point Finding f’’(𝒙) f’’(π‘₯)=6 𝑑(π‘₯^2 βˆ’ 2π‘₯ + 1)/𝑑π‘₯ f’’(π‘₯)=6(2π‘₯βˆ’2+0) f’’(π‘₯)=12(π‘₯βˆ’1) Putting 𝒙=𝟏 f’’(1)=12(1βˆ’1) = 12 Γ— 0 = 0 Since f’’(1) = 0 Hence, π‘₯=1 is neither point of Maxima nor point of Minima ∴ 𝒙=𝟏 is Point of Inflexion.

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