Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12     1. Chapter 6 Class 12 Application of Derivatives
2. Serial order wise
3. Examples

Transcript

Example 29 (Method 1) Find all points of local maxima and local minima of the function f given by 𝑓 (𝑥)=𝑥3 – 3𝑥 + 3. 𝑓(𝑥)=𝑥3 – 3𝑥+3 Finding 𝒇^′ (𝒙) 𝑓′(𝑥)= 3𝑥^2 – 3+0 𝑓′(𝑥)= 3(𝑥^2−1) Putting 𝒇′(𝒙)= 0 3(𝑥^2−1)=0 𝑥^2−1=0 (𝑥−1)(𝑥+1)=0 Thus 𝑥 = 1 , –1 are only critical points Finding maximum & minimum value of 𝑓(𝑥)=𝑥3 – 3𝑥+3 Minimum value 𝑓(1)=(1)3 –3(1)+3= 1 – 3 + 3 = 1 Maximum value 𝑓(−1)=(−1)3 –3(−1)+3= –1 +3 + 3 = 5 Example 29 (Method 2) Find all points of local maxima and local minima of the function f given by 𝑓 (𝑥) = 𝑥3 – 3𝑥 + 3. 𝑓(𝑥)=𝑥3 – 3𝑥+3 Finding 𝒇^′ (𝒙) 𝑓′(𝑥)= 3𝑥^2 – 3+0 𝑓′(𝑥)= 3(𝑥^2−1) Putting 𝒇′(𝒙)= 0 3(𝑥^2−1)=0 𝑥^2−1=0 (𝑥−1)(𝑥+1)=0 So, x = 1 & x = −1 Finding 𝒇′′(𝒙) 𝑓^′ (𝑥)=3(𝑥^2−1) 𝑓^′′ (𝑥)=3 𝑑(𝑥^2 − 1)/𝑑𝑥 𝑓^′′ (𝑥)=3(2𝑥−0) 𝑓^′′ (𝑥)=6𝑥 Putting 𝒙=−𝟏 𝑓^′′ (𝑥) = 6(−1) = −6<0 𝑥 = –1 is point of local maxima Putting 𝒙=𝟏 𝑓^′′ (𝑥) = 6(1) = 6>0 𝑥 = 1 is point of local minima Finding maximum & minimum value of 𝑓(𝑥)=𝑥3 – 3𝑥+3 Minimum value at x = 1 𝑓(1)=(1)3 –3(1)+3= 1 – 3 + 3 = 1 Maximum value at x = −1 𝑓(−1)=(−1)3 –3(−1)+3= –1 + 3 + 3 = 5

Examples 