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

Slide7.JPG
Slide8.JPG Slide9.JPG Slide10.JPG

  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise
Ask Download

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 Step 1: Finding f’﷐𝑥﷯ 𝑓’﷐𝑥﷯=﷐𝑑﷐2𝑥3 − 6𝑥2 + 6𝑥 + 5﷯﷮𝑑𝑥﷯ 𝑓’﷐𝑥﷯=6﷐𝑥﷮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﷯ Step 2: Putting f’﷐𝑥﷯=0 6﷐﷐𝑥−1﷯﷮2﷯=0 ﷐﷐𝑥−1﷯﷮2﷯=0 𝑥−1=0 So, 𝑥=1 Step 3: 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 Step 1: Finding f’﷐𝑥﷯ 𝑓’﷐𝑥﷯=﷐𝑑﷐2𝑥3 − 6𝑥2+ 6𝑥 + 5 ﷯﷮𝑑𝑥﷯ 𝑓’﷐𝑥﷯=6﷐𝑥﷮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﷯ Step 2: Putting f’﷐𝑥﷯=0 6﷐﷐𝑥−1﷯﷮2﷯=0 ﷐﷐𝑥−1﷯﷮2﷯=0 𝑥−1=0 𝑥=1 Step 3: Finding f’’﷐𝑥﷯ f’﷐𝑥﷯=6﷐﷐𝑥−1﷯﷮2﷯ 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 ∴ Second derivative test fails ⇒ 𝑥=1 is neither point of maxima nor point of minima Thus, 𝒙=𝟏 is point of inflexion

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.