     1. Chapter 6 Class 12 Application of Derivatives (Term 1)
2. Serial order wise
3. Examples

Transcript

Example 33 (Method 1) Find all the points of local maxima and local minima of the function f given by 𝑓(𝑥)=2𝑥3 –6𝑥2+6𝑥+5. 𝑓(𝑥)=2𝑥3 –6𝑥2+6𝑥+5 Finding f’(𝒙) 𝑓’(𝑥)=𝑑(2𝑥3 − 6𝑥2 + 6𝑥 + 5)/𝑑𝑥 𝑓’(𝑥)=6𝑥^2−12𝑥+6+0 𝑓’(𝑥)=6(𝑥^2−2𝑥+1) 𝑓’(𝑥)=6((𝑥)^2+(1)^2−2(𝑥)(1)) 𝑓’(𝑥)=𝟔(𝒙−𝟏)^𝟐 Putting f’(𝒙)=𝟎 6(𝑥−1)^2=0 (𝑥−1)^2=0 𝑥−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(𝑥)=2𝑥3 –6𝑥2+6𝑥+5. 𝑓(𝑥)=2𝑥3 –6𝑥2+6𝑥+5 Finding f’(𝒙) 𝑓’(𝑥)=𝑑(2𝑥3 − 6𝑥2 + 6𝑥 + 5" " )/𝑑𝑥 𝑓’(𝑥)=6𝑥^2−12𝑥+6++0 𝑓’(𝑥)=6(𝑥^2−2𝑥+1) 𝑓’(𝑥)=6((𝑥)^2+(1)^2−2(𝑥)(1)) 𝑓’(𝑥)=𝟔(𝒙−𝟏)^𝟐 Putting f’(𝒙)=𝟎 6(𝑥−1)^2=0 (𝑥−1)^2=0 So, 𝒙=𝟏 is the only critical point Finding f’’(𝒙) f’’(𝑥)=6.(𝑑(𝑥 − 1)^2)/𝑑𝑥 f’’(𝑥)=6 × 2(𝑥−1) f’’(𝑥) = 12 (𝑥−1) Putting 𝒙=𝟏 f’’(𝑥)=12(1−1) = 0 Since f’’(1) = 0 Hence, 𝑥=1 is neither point of Maxima nor point of Minima ∴ 𝒙=𝟏 is Point of Inflexion. 