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Example 32 - Find local maximum and local minimum values - Local maxima and minima

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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise
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Example 32 (Method 1) Find local maximum and local minimum values of the function f given by f (𝑥)=3𝑥4 + 4𝑥3 – 12𝑥2 + 12 f (𝑥)=3𝑥4 + 4𝑥3 – 12𝑥2 + 12 Step 1: Finding f’ (𝑥) f’ (𝑥)=﷐𝑑﷐3𝑥4 + 4𝑥3 – 12𝑥2 + 12 ﷯﷮𝑑𝑥﷯ f’ (𝑥)=12﷐𝑥﷮3ï·¯+12﷐𝑥﷮2ï·¯ – 24𝑥 + 0 f’ (𝑥)=12𝑥3 +12𝑥2 – 24𝑥 f’ (𝑥)=12﷐﷐𝑥﷮3ï·¯+﷐𝑥﷮2﷯−2𝑥﷯ f’ ﷐𝑥﷯=12𝑥 ﷐﷐𝑥﷮2ï·¯+2𝑥−𝑥−2ï·¯ f’ ﷐𝑥﷯=12𝑥 ﷐𝑥﷐𝑥+2﷯−1﷐𝑥+2﷯﷯ f’ ﷐𝑥﷯=12𝑥 ﷐𝑥−1﷯﷐𝑥+2ï·¯ Step 2: Putting f’ ﷐𝑥﷯=0 12𝑥 ﷐𝑥−1﷯﷐𝑥+2ï·¯=0 𝑥 ﷐𝑥−1﷯﷐𝑥+2ï·¯=0 So, 𝑥=0 , 1 & –2 are critical points Step 3: Step 4: Finding local minimum and maximum value Local maximum value of f (𝑥) at 𝑥=0 f (0)=3﷐0ï·¯4 + 4﷐0ï·¯3 – 12﷐0ï·¯2 + 12 = 0 + 0 – 0 + 12 = 12 Local minimum value of f (𝑥) at 𝑥=1 f (1)=3﷐1ï·¯4 + 4﷐1ï·¯3 – 12﷐1ï·¯2 + 12 = 3 + 4 – 12 + 12= 7 Local Minimum value of f (𝑥) at 𝑥=−2 f (−2)=3﷐−2ï·¯4 + 4﷐−2ï·¯3 – 12﷐−2ï·¯2 + 12 = 48 – 32 – 48 + 12 = – 20 Example 32(Method 2) Find local maximum and local minimum values of the function f given by f (𝑥)=3𝑥4 + 4𝑥3 – 12𝑥2 + 12 f (𝑥)=3𝑥4 + 4𝑥3 – 12𝑥2 + 12 Step 1: Finding f’ (𝑥) f’ (𝑥)=﷐𝑑﷐3𝑥4 + 4𝑥3 – 12𝑥2 + 12 ﷯﷮𝑑𝑥﷯ f’ (𝑥)=12﷐𝑥﷮3﷯+12﷐𝑥﷮2﷯ – 24𝑥 + 0 f’ (𝑥)=12𝑥3+12𝑥3 – 24𝑥 f’ (𝑥)=12﷐﷐𝑥﷮3﷯+﷐𝑥﷮2﷯−2𝑥﷯ f’ ﷐𝑥﷯=12𝑥 ﷐﷐𝑥﷮2﷯+𝑥−2﷯ f’ ﷐𝑥﷯=12𝑥 ﷐﷐𝑥﷮2﷯+2𝑥−𝑥−2﷯ f’ ﷐𝑥﷯=12𝑥 ﷐𝑥﷐𝑥+2﷯−1﷐𝑥+2﷯﷯ f’ ﷐𝑥﷯=12𝑥 ﷐𝑥−1﷯﷐𝑥+2﷯ Step 2: Putting f’ ﷐𝑥﷯=0 12𝑥 ﷐𝑥−1﷯﷐𝑥+2﷯=0 𝑥 ﷐𝑥−1﷯﷐𝑥+2﷯=0 So, 𝑥=0,𝑥=1,& 𝑥=−1 Step 3: Finding f’’﷐𝑥﷯ f ’﷐𝑥﷯=12﷐﷐𝑥﷮3﷯+﷐𝑥﷮2﷯−2𝑥﷯ f ’’﷐𝑥﷯=﷐12𝑑﷐﷐𝑥﷮3﷯ + ﷐𝑥﷮2﷯ − 2𝑥﷯﷮𝑑𝑥﷯ f ’’﷐𝑥﷯=12﷐3﷐𝑥﷮2﷯+2𝑥−2﷯ At 𝒙=𝟎 f ’’﷐0﷯=12﷐3﷐﷐0﷯﷮2﷯+2﷐0﷯−2﷯= 32 ﷐0+0 −2﷯= – 64 < 0 Since f’’﷐𝑥﷯<0 at 𝑥=0 ⇒ 𝑥 is point of local maxima Thus, f﷐𝑥﷯ is maximum at 𝑥=0 At 𝒙=𝟏 f’﷐1﷯=12﷐3﷐﷐1﷯﷮2﷯+2﷐1﷯−2﷯= 12 ﷐3+2−2﷯ = 36 > 0 Since f’’﷐𝑥﷯>0 at 𝑥=1 ⇒ 𝑥 is point of local minima Thus, f﷐𝑥﷯ is minimum at 𝑥=1 At 𝒙=−𝟐 f’﷐−2﷯=12﷐3﷐﷐−2﷯﷮2﷯+2﷐−2﷯−2﷯= 12 ﷐3﷐4﷯−4−2﷯= 72 > 0 Since f’’﷐𝑥﷯>0 at 𝑥=−2 ⇒ 𝑥 is point of local minima Thus, f﷐𝑥﷯ is minimum at 𝑥=−2 Step 4: Finding local minimum and maximum value Local maximum value of f (𝑥) at 𝑥=0 f (0)=3﷐0﷯4 + 4﷐0﷯3 – 12﷐0﷯2 + 12 = 0 + 0 – 0 + 12 = 12 Local minimum value of f (𝑥) at 𝑥=1 f (1)=3﷐1﷯4 + 4﷐1﷯3 – 12﷐1﷯2 + 12 = 3 + 4 – 12 + 12 = 7 Local Minimum value of f (𝑥) at 𝑥=−2 f (−2)=3﷐−2﷯4 + 4﷐−2﷯3 – 12﷐−2﷯2 + 12 = 48 – 32 – 48 + 12 = – 20

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