Check sibling questions

Example 33 - Find all points of local maxima, minima - NCERT

Example 33 - Chapter 6 Class 12 Application of Derivatives - Part 2

Example 33 - Chapter 6 Class 12 Application of Derivatives - Part 3

Example 33 - Chapter 6 Class 12 Application of Derivatives - Part 4
Example 33 - Chapter 6 Class 12 Application of Derivatives - Part 5


Transcript

Example 33 (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’(π‘₯)=𝑑(2π‘₯3 βˆ’ 6π‘₯2 + 6π‘₯ + 5)/𝑑π‘₯ 𝑓’(π‘₯)=6π‘₯^2βˆ’12π‘₯+6+0 𝑓’(π‘₯)=6(π‘₯^2βˆ’2π‘₯+1) 𝑓’(π‘₯)=6((π‘₯)^2+(1)^2βˆ’2(π‘₯)(1)) 𝑓’(π‘₯)=πŸ”(π’™βˆ’πŸ)^𝟐 Putting f’(𝒙)=𝟎 6(π‘₯βˆ’1)^2=0 (π‘₯βˆ’1)^2=0 π‘₯βˆ’1=0 So, 𝒙=𝟏 is only critical point Hence 𝒙=𝟏 is point of inflexion Example 33 (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’(π‘₯)=𝑑(2π‘₯3 βˆ’ 6π‘₯2 + 6π‘₯ + 5" " )/𝑑π‘₯ 𝑓’(π‘₯)=6π‘₯^2βˆ’12π‘₯+6++0 𝑓’(π‘₯)=6(π‘₯^2βˆ’2π‘₯+1) 𝑓’(π‘₯)=6((π‘₯)^2+(1)^2βˆ’2(π‘₯)(1)) 𝑓’(π‘₯)=πŸ”(π’™βˆ’πŸ)^𝟐 Putting f’(𝒙)=𝟎 6(π‘₯βˆ’1)^2=0 (π‘₯βˆ’1)^2=0 So, 𝒙=𝟏 is the only critical point Finding f’’(𝒙) f’’(π‘₯)=6.(𝑑(π‘₯ βˆ’ 1)^2)/𝑑π‘₯ f’’(π‘₯)=6 Γ— 2(π‘₯βˆ’1) f’’(π‘₯) = 12 (π‘₯βˆ’1) Putting 𝒙=𝟏 f’’(π‘₯)=12(1βˆ’1) = 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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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.