Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives

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Ex 6.5, 3 Find the local maxima and local minima, if any, of the following functions. Find also the local maximum & the local minimum values, as the case may be: (i) f (𝑥)=𝑥2 f(𝑥)=𝑥^2 Finding f’(x) f’(x) = 2x Putting f’(x) = 0 2x = 0 x = 0 Finding f’’(x) f’(x) = 2x Differentiating again f’’(x) = 2 Since f’’(x) > 0 for x = 0 So, f(x) is minimum at x = 0 Minimum value of f(x) at x = 0 f(x) = x2 f(0) = 02 = 0 So, Minimum value of f(x) = 0 There is no maximum value Ex 6.5, 3 Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be: (ii) 𝑔(𝑥)=𝑥3 –3𝑥 𝑔(𝑥)=𝑥3 –3𝑥 Finding g’(𝒙) g’(𝑥)=𝑑(𝑥^3 − 3𝑥)/𝑑𝑥 g’(𝑥)=3𝑥^2−3 Putting g’(𝒙)=𝟎 3𝑥^2−3=0 3𝑥^2=3 𝑥^2=3/3 𝑥^2=1 𝑥=±1 So, x = 1 & x = –1 Finding g’’(𝒙) g’(𝑥)=3𝑥^2−3 g’’(𝑥)=𝑑(3𝑥^2−3)/𝑑𝑥 = 6𝑥−0 = 6𝑥 Putting 𝒙=𝟏 in g’’(x) g’’(1)=6(1)= 6 > 0 Thus, g’’(𝑥)>0 when 𝑥=1 ⇒ 𝑥=1 is point of local minima & g(𝑥) is minimum at 𝑥=1 Local minimum value g(𝑥)=𝑥^3−3𝑥 g(1)=(1)^3−3(1) =1−3 =−𝟐 Putting 𝒙=−𝟏 in g’’(x) g’’(−1)=6(−1)= –6 < 0 Thus, g’’(𝑥)<0 when 𝑥=−1 ⇒ 𝑥=−1 is point of local maxima & g(𝑥) is maximum at 𝑥=−1 Local minimum value g(𝑥)=𝑥^3−3𝑥 g(−1)=(−1)^3−3(−1) =−1+3 =𝟐