Example 30 - Chapter 6 Class 12 Application of Derivatives
Last updated at May 29, 2018 by Teachoo
Transcript
Example 30 (Method 1 ) Find all the points of local maxima and local minima of the unction f given by f 𝑥=2𝑥3 –6𝑥2+6𝑥+5. f (𝑥)=2𝑥3 –6𝑥2+ 6𝑥+5 Step 1: f’ (𝑥) f ′(𝑥)= 𝑑2𝑥3 – 6𝑥2+ 6𝑥 + 5𝑑𝑥 f ′(𝑥)=6𝑥2 –12𝑥+6+0 f ′(𝑥)=6𝑥2−2𝑥+1 Step 2: Putting f ′(𝑥)= 0 6𝑥2−2𝑥+1=0 𝑥2−2𝑥+1=0 𝑥2+12−2𝑥1=0 𝑥−12=0 𝑥−1𝑥−1=0 So, 𝑥=1 is only critical point Step 3: Hence 𝒙=𝟏 is point of inflexion Example 30 (Method 2) Find all the points of local maxima and local minima of the unction f given by f 𝑥=2𝑥3 –6𝑥2+6𝑥+5. f (𝑥)=2𝑥3 –6𝑥2+ 6𝑥+5 Step 1: Finding f’ (𝑥) f ′(𝑥)= 𝑑2𝑥3 – 6𝑥2+ 6𝑥 + 5𝑑𝑥 f ′(𝑥)=6𝑥3 –12𝑥+6+0 f ′(𝑥)=6𝑥2−2𝑥+1 Step 2: Putting f ′(𝑥)= 0 6𝑥2−2𝑥+1=0 𝑥2−2𝑥+1=0 𝑥2+12−2𝑥1=0 𝑥−12=0 𝑥−1𝑥−1=0 So, 𝑥=1 is only critical point Step 3: Finding f’’𝑥 f’𝑥=6𝑥2−2𝑥+1 f’’𝑥=6𝑑𝑥2 − 2𝑥 + 1𝑑𝑥 f’’𝑥=62𝑥−2+0 f’’𝑥=12𝑥−1 Putting 𝑥=1 f’’1=121−1 f’’𝑥=12𝑥−1 At x = 1, f’’𝑥 =120= 0 Thus, second derivative test fail Hence, 𝑥=1 is neither point of Maxima nor point of Minima ∴ 𝒙=𝟏 is point of inflexion.
Examples
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Example 4
Example 5
Example 6
Example 7
Example 8
Example 9
Example 10
Example 11
Example 12
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Example 15
Example 16
Example 17
Example 18
Example 19
Example 20
Example 21
Example 22
Example 23
Example 24
Example 25
Example 26
Example 27
Example 28
Example 29
Example 30 You are here
Example 31
Example 32
Example 33
Example 34
Example 35 Important
Example 36
Example 37 Important
Example 38 Important
Example 39
Example 40 Important
Example 41
Example 42
Example 43
Example 44
Example 45
Example 46 Important
Example 47 Important
Example 48
Example 49
Example 50
Example 51
Chapter 6 Class 12 Application of Derivatives
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About the Author
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.