Example 31 - Find local minimum value of f(x) = 3 + |x| - Local maxima and minima

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  1. Chapter 6 Class 12 Application of Derivatives
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Example 31 (Method 1) Find local minimum value of the function f given by f (𝑥) = 3 + |𝑥|, 𝑥 ∈ R. f (𝑥) = 3 + |𝑥| 𝑥 ∈ R Since Value of ﷐𝑥﷯≥0 So, Minimum value of ﷐𝑥﷯=0 ⇒ Minimum value of f (𝑥) = 3 + Minimum value of ﷐𝑥﷯ = 3 + 0 = 3 Hence minimum value of f (𝑥) = 3 We cannot find any maximum value. Example 31 (Method 2) Find local minimum value of the function f given by f (𝑥) = 3 + |𝑥|, 𝑥 ∈ R. f (𝑥) = 3 + |𝑥| We know that ﷐𝑥﷯=﷐﷐−&𝑥, 𝑥<0﷮&𝑥, 𝑥≥0﷯﷯ Hence, f (𝑥) =﷐﷐3−&𝑥, 𝑥<0﷮3+𝑥, 𝑥≥0﷯﷯ 𝑥 = 0 is only critical point We check sign of f’ (𝑥) at 𝑥 = 0 Since sign of f’(x) changes from negative to positive, x = 0 is point of Maxima & f (𝑥) is Minimum at 𝑥 = 0 Minimum value of f (𝑥) f (0) = 3 + |0|= 3 There’s no Maximum value of f (𝒙) Example 31 (Method 3) Find local minimum value of the function f given by f (𝑥) = 3 + |𝑥|, 𝑥 ∈ R. f (𝑥) = 3 + |𝑥| From graph, f (𝑥) is Minimum at 𝑥 = 0 So, Minimum value of f (𝑥) = f (0) = 3

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