



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.
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