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Ex 6.5, 3 - Find local maxima, local minima of (i) f(x) = x2 - Ex 6.5

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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise
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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 and the local minimum values, as the case may be: (i) f (𝑥)=𝑥2 f 𝑥﷯= 𝑥﷮2﷯ First we plot the graph of 𝑥2 Note that :- At 𝑥 = 0 , f 0﷯=0 Also, f 𝑥﷯>0 for all 𝑥 , 𝑥 ∈ R expect 0 Hence, x = 0 is point of minima of f (x) So, Minimum value of f 𝑥﷯=0 at 𝑥 = 0 Since f 𝑥﷯>0 for 𝑥 ∈ R expect 0 So, we can not find a maximum value e.g. f 100﷯ = 100﷯﷮2﷯= 10000 f 1000﷯= 1000﷯﷮2﷯= 1000000 Hence , we cannot find a maximum value of f 𝑥﷯ on , 𝑥 ∈ R Hence no point of maximum value of f on R Ex 6.5,3 (Method 1) 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 Step 1: Finding g’ 𝑥﷯ g’ 𝑥﷯= 𝑑 𝑥﷮3﷯−3𝑥﷯﷮𝑑𝑥﷯ g’ 𝑥﷯=3 𝑥﷮2﷯−3 Step 2: Putting g’ 𝑥﷯=0 3 𝑥﷮2﷯−3=0 3 𝑥﷮2﷯=3 𝑥﷮2﷯= 3﷮3﷯ 𝑥﷮2﷯=1 𝑥=±1 So, x = 1 & x = –1 Step 3: First derivative test Thus, ⇒ 𝑥=1 is point of local minima ⇒ g 𝑥﷯ Is minimum at 𝑥=1 & g’ 𝑥﷯ changes sign from +﷯𝑣𝑒 to −﷯𝑣𝑒 when 𝑥=−1 ⇒ 𝑥=−1 is point of local maxima ⇒ g 𝑥﷯ is maximum at 𝑥=−1 Local minimum value of g 𝑥﷯= 𝑥﷮3﷯−3𝑥 at x = 1 g 1﷯= 1﷯﷮3﷯−3 1﷯=1−3=−𝟐 Local maximum value of g 𝑥﷯= 𝑥﷮3﷯−3𝑥 at x = –1 g(−1)= −1﷯﷮3﷯−3 −1﷯=−1+3 =𝟐 Ex 6.5,3 (Method 2) 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 Step 1: Finding g’ 𝑥﷯ g’ 𝑥﷯= 𝑑 𝑥﷮3﷯−3𝑥﷯﷮𝑑𝑥﷯ g’ 𝑥﷯=3 𝑥﷮2﷯−3 Step 2: Putting g’ 𝑥﷯=0 3 𝑥﷮2﷯−3=0 3 𝑥﷮2﷯=3 𝑥﷮2﷯= 3﷮3﷯ 𝑥﷮2﷯=1 𝑥=±1 So, x = 1 & x = –1 Step 3: Finding g’’ 𝑥﷯ g’ 𝑥﷯=3 𝑥﷮2﷯−3 g’’ 𝑥﷯= 𝑑 3 𝑥﷮2﷯−3﷯﷮𝑑𝑥﷯ = 6𝑥−0 = 6𝑥 Ex 6.5,3 (Method 1) 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: (iii) ℎ(𝑥)=sin⁡𝑥+cos⁡𝑥, 0<𝑥< 𝜋﷮2﷯ ℎ(𝑥)=sin⁡𝑥+cos⁡𝑥, 0<𝑥< 𝜋﷮2﷯ Step 1: Finding ℎ′(𝑥) ℎ′(𝑥)= 𝑑 sin﷮𝑥﷯+ cos⁡𝑥 ﷯﷮𝑑𝑥﷯ ℎ﷮′﷯ 𝑥﷯= cos﷮𝑥﷯− sin﷮𝑥﷯ Step 2: Putting ℎ(𝑥)=0 cos﷮𝑥﷯−𝑠𝑖𝑛𝑥=0 cos﷮𝑥=𝑠𝑖𝑛 𝑥﷯ 1 = sin﷮𝑥﷯﷮ cos﷮𝑥﷯﷯ 1 = tan 𝑥 tan 𝑥=1 ∴ 𝑥=45°= 𝜋﷮4﷯ Step 3: Since sign of f’(x) changes from positive to negative, x = 𝜋﷮4﷯ is Maxima Finding maximum value f has Maximum value at 𝑥= π﷮4﷯ f 𝑥﷯=𝑠𝑖𝑛𝑥+𝑐𝑜𝑠𝑥 f π﷮4﷯﷯=𝑠𝑖𝑛 π﷮4﷯﷯+𝑐𝑜𝑠 π﷮4﷯﷯ = 1﷮ ﷮2﷯﷯+ 1﷮ ﷮2﷯﷯ = 2﷮ ﷮2﷯﷯ = ﷮𝟐﷯ Ex 6.5,3 (Method 2) 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: (iii) ℎ(𝑥)=sin⁡𝑥+cos⁡𝑥, 0<𝑥< 𝜋﷮2﷯ ℎ(𝑥)=sin⁡𝑥+cos⁡𝑥, 0<𝑥< 𝜋﷮2﷯ Step 1: Finding ℎ′(𝑥) ℎ′(𝑥)= 𝑑 sin﷮𝑥﷯+ cos⁡𝑥 ﷯﷮𝑑𝑥﷯ ℎ﷮′﷯ 𝑥﷯= cos﷮𝑥﷯− sin﷮𝑥﷯ Step 2: Putting ℎ(𝑥)=0 cos﷮𝑥﷯−𝑠𝑖𝑛𝑥=0 cos﷮𝑥=𝑠𝑖𝑛 𝑥﷯ 1 = sin﷮𝑥﷯﷮ cos﷮𝑥﷯﷯ 1 = tan 𝑥 tan 𝑥=1 ∴ 𝑥=45°= 𝜋﷮4﷯ Step 3: Finding h’’ 𝑥﷯ h’ 𝑥﷯=𝑐𝑜𝑠𝑥−𝑠𝑖𝑛𝑥 h’’ 𝑥﷯=−𝑠𝑖𝑛𝑥−𝑐𝑜𝑠𝑥 Putting 𝑥= π﷮4﷯ h’’ π﷮4﷯﷯=−sin π﷮4﷯﷯−𝑐𝑜𝑠 π﷮4﷯﷯ = – 1﷮ ﷮2﷯﷯− 1﷮ ﷮2﷯﷯ = −2﷮ ﷮2﷯﷯ = – ﷮2﷯ Since h’’ 𝑥﷯<0 when 𝑥= π﷮4﷯ ⇒𝑥= π﷮4﷯ is point of Local Maxima f has Maximum value at 𝒙= 𝝅﷮𝟒﷯ f 𝑥﷯=𝑠𝑖𝑛𝑥+𝑐𝑜𝑠𝑥 f π﷮4﷯﷯=𝑠𝑖𝑛 π﷮4﷯﷯+𝑐𝑜𝑠 π﷮4﷯﷯ = 1﷮ ﷮2﷯﷯+ 1﷮ ﷮2﷯﷯ = 2﷮ ﷮2﷯﷯ = ﷮𝟐﷯ 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: (iv) f (𝑥)=sin⁡𝑥 –cos⁡𝑥, 0<𝑥<2 𝜋 f (𝑥)=sin⁡𝑥 –cos⁡𝑥, 0<𝑥<2 𝜋 Finding f’ 𝑥﷯ f’ 𝑥﷯= cos﷮𝑥﷯− − sin﷮𝑥﷯﷯ f’ 𝑥﷯= cos﷮𝑥﷯+ sin﷮𝑥﷯ Putting f’ 𝑥﷯=0 cos﷮𝑥﷯+ sin﷮𝑥﷯ = 0 cos﷮𝑥﷯=− sin﷮𝑥﷯ 1= − sin﷮𝑥﷯﷮ cos﷮𝑥﷯﷯ − sin﷮𝑥﷯﷮ cos﷮𝑥﷯﷯=1 – tan 𝑥=1 tan 𝑥=−1 Since 0 < 𝑥 < 2π & tan 𝑥 is negative tan θ lies in either (ii) or (iv) quadrant So, value of 𝑥 is 𝑥= 3𝜋﷮4﷯𝑜𝑟 7𝜋﷮4﷯ Now finding f’’ 𝑥﷯ f’’ 𝑥﷯= 𝑑 cos﷮𝑥﷯ + sin﷮𝑥﷯﷯﷮𝑑𝑥﷯ f’’ 𝑥﷯=− sin﷮𝑥﷯+ cos﷮𝑥﷯ Putting 𝒙 = 𝟑𝝅﷮𝟒﷯ f’’ 3𝜋﷮4﷯﷯=−𝑠𝑖𝑛 3𝜋﷮4﷯﷯+𝑐𝑜𝑠 3𝜋﷮4﷯﷯ =−𝑠𝑖𝑛 𝜋− 𝜋﷮4﷯﷯+𝑐𝑜𝑠 𝜋− 𝜋﷮4﷯﷯ =−𝑠𝑖𝑛 𝜋﷮4﷯﷯+ −𝑐𝑜𝑠 𝜋﷮4﷯﷯ = −1﷮ ﷮2﷯﷯− 1﷮ ﷮2﷯﷯ = −2﷮ ﷮2﷯﷯ =− ﷮2﷯ < 0 Hence f’’ 𝑥﷯<0 when 𝑥 = 3𝜋﷮4﷯ Thus 𝑥 = 3𝜋﷮4﷯ is point of local maxima ∴ f 𝑥﷯ is maximum value at 𝑥 = 3𝜋﷮4﷯ The local maximum value is f 𝑥﷯= sin﷮𝑥﷯− cos﷮𝑥﷯ f 3𝜋﷮4﷯﷯=𝑠𝑖𝑛 3𝜋﷮4﷯﷯−𝑐𝑜𝑠 3𝜋﷮4﷯﷯ =𝑠𝑖𝑛 𝜋− 𝜋﷮4﷯﷯−𝑐𝑜𝑠 𝜋− 𝜋﷮4﷯﷯ =𝑠𝑖𝑛 𝜋﷮4﷯﷯− −𝑐𝑜𝑠 𝜋﷮4﷯﷯ =𝑠𝑖𝑛 𝜋﷮4﷯+𝑐𝑜𝑠 𝜋﷮4﷯ = 1﷮ ﷮2﷯﷯+ 1﷮ ﷮2﷯﷯ = 2﷮ ﷮2﷯﷯ = ﷮2﷯ Now, for 𝒙 = 𝟕𝝅﷮𝟒﷯ f’’ 𝑥﷯=− sin﷮𝑥﷯+ cos﷮𝑥﷯ Putting 𝑥 = 7𝜋﷮4﷯ f’’ 7𝜋﷮4﷯﷯=− sin﷮ 7𝜋﷮4﷯﷯﷯+ cos﷮ 7𝜋﷮4﷯﷯﷯ f’’ 7𝜋﷮4﷯﷯=−𝑠𝑖𝑛 2𝜋− 𝜋﷮4﷯﷯+𝑐𝑜𝑠 2𝜋− 𝜋﷮4﷯﷯ =− −𝑠𝑖𝑛 𝜋﷮4﷯﷯﷯+𝑐𝑜𝑠 𝜋﷮4﷯﷯ =𝑠𝑖𝑛 𝜋﷮4﷯+𝑐𝑜𝑠 𝜋﷮4﷯ = 1﷮ ﷮2﷯﷯ + 1﷮ ﷮2﷯﷯ = 2﷮ ﷮2﷯﷯ = ﷮2﷯ > 0 f’’ 𝑥﷯>0 when 𝑥 = 7𝜋﷮4﷯ Thus 𝑥 = 7𝜋﷮4﷯ is point of local minima f 𝑥﷯ has minimum value at 𝑥 = 7𝜋﷮4﷯ Local minimum value is f 𝑥﷯=𝑠𝑖𝑛𝑥−𝑐𝑜𝑠𝑥 f 7𝜋﷮4﷯﷯=𝑠𝑖𝑛 7𝜋﷮4﷯﷯−𝑐𝑜𝑠 7𝜋﷮4﷯﷯ =𝑠𝑖𝑛 2𝜋− 𝜋﷮4﷯﷯−𝑐𝑜𝑠 2𝜋− 𝜋﷮4﷯﷯ =−𝑠𝑖𝑛 𝜋﷮4﷯﷯−𝑐𝑜𝑠 𝜋﷮4﷯﷯ = −1﷮ ﷮2﷯﷯ − 1﷮ ﷮2﷯﷯ = −2﷮ ﷮2﷯﷯ =− ﷮2﷯ Thus, f 𝑥﷯ is maximum at x = 𝟑𝝅﷮𝟒﷯ and maximum value is ﷮𝟐﷯ & f 𝑥﷯ is minimum at x = 𝟕𝝅﷮𝟒﷯ and maximum value is – ﷮𝟐﷯ Ex 6.5,3(Method 1) 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: (v) 𝑓 𝑥﷯=𝑥3 –6𝑥2+9𝑥+15 𝑓 𝑥﷯=𝑥3 –6𝑥2+9𝑥+15 Finding f’ 𝑥﷯ f’ 𝑥﷯= 𝑑 𝑥3 – 6𝑥2+ 9𝑥 + 15 ﷯﷮𝑑𝑥﷯ f’ 𝑥﷯=3 𝑥﷮2﷯−12𝑥+9 f’ 𝑥﷯=3 𝑥﷮2﷯−4𝑥+3﷯ Putting f’ 𝑥﷯=0 3 𝑥﷮2﷯−4𝑥+3﷯=0 𝑥﷮2﷯−4𝑥+3=0 𝑥﷮2﷯−3𝑥−𝑥+3=0 𝑥 𝑥−3﷯−1 𝑥−3﷯=0 𝑥−1﷯ 𝑥−3﷯=0 So, x = 1 & x = 3 Step 3: Using first derivative test at x = 1 & x = −3 Thus, ⇒ 𝑥 = 1 is point of local maxima f 𝑥﷯ has maximum value at 𝑥 = 1 And, 𝑥 = 3 is point of local minima ⇒ f 𝑥﷯ has minimum value at 𝑥 = 3 Maximum value of f 𝑥﷯ at 𝒙 = 1 f 𝑥﷯= 𝑥﷮3﷯−6 𝑥﷮2﷯+9𝑥+15 f 𝑥﷯= 1﷯﷮3﷯+6 1﷯﷮2﷯+9 1﷯+15 = 1 – 6 + 9 + 15 = 4 + 15 = 19 Minimum value of f 𝑥﷯ at 𝒙 = 3 f 𝑥﷯= 𝑥﷮3﷯−6 𝑥﷮2﷯+9𝑥+15 f 𝑥﷯= 3﷯﷮3﷯−6 3﷯﷮2﷯+9 3﷯+15 = 27 – 54 + 27 + 15 = 15 Ex 6.5,3(Method 2) 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: (v) 𝑓 𝑥﷯=𝑥3 –6𝑥2+9𝑥+15 𝑓 𝑥﷯=𝑥3 –6𝑥2+9𝑥+15 Finding f’ 𝑥﷯ f’ 𝑥﷯= 𝑑 𝑥3 – 6𝑥2+ 9𝑥 + 15 ﷯﷮𝑑𝑥﷯ f’ 𝑥﷯=3 𝑥﷮2﷯−12𝑥+9 f’ 𝑥﷯=3 𝑥﷮2﷯−4𝑥+3﷯ Putting f’ 𝑥﷯=0 3 𝑥﷮2﷯−4𝑥+3﷯=0 𝑥﷮2﷯−4𝑥+3=0 𝑥﷮2﷯−3𝑥−𝑥+3=0 𝑥 𝑥−3﷯−1 𝑥−3﷯=0 𝑥−1﷯ 𝑥−3﷯=0 So, x = 1 & x = 3 Step 3: Finding f’’ 𝑥﷯ f’ 𝑥﷯=3 𝑥﷮2﷯−4𝑥+3﷯ f’’ 𝑥﷯= 𝑑 3 𝑥﷮2﷯−4𝑥+3﷯﷯﷮𝑑𝑥﷯ = 3 𝑑 𝑥﷮2﷯−4𝑥+3﷯﷮𝑑𝑥﷯ = 3 2𝑥−4+0﷯ = 6𝑥−12 ∴ f’’ 𝑥﷯=6𝑥−12 Ex 6.5,3 (Method 1) 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: (vi) g (𝑥) = 𝑥﷮2﷯ + 2﷮𝑥﷯ , 𝑥 > 0 g (𝑥) = 𝑥﷮2﷯ + 2﷮𝑥﷯ , 𝑥 > 0 Finding g’ 𝑥﷯ g’ 𝑥﷯= 𝑑﷮𝑑𝑥﷯ 𝑥﷮2﷯+ 2﷮𝑥﷯﷯ = 𝑑﷮𝑑𝑥﷯ 𝑥﷮2﷯﷯+ 𝑑﷮𝑑𝑥﷯ 2 𝑥﷮−1﷯﷯ = 1﷮2﷯−2 𝑥﷮−2﷯ = 1﷮2﷯− 2﷮ 𝑥﷮2﷯﷯ Putting g’ 𝑥﷯=0 1﷮2﷯− 2﷮ 𝑥﷮2﷯﷯=0 𝑥﷮2﷯− 4﷮2 𝑥﷮2﷯﷯=0 𝑥﷮2﷯−4=0 ×2 𝑥﷮2﷯ 𝑥−2﷯ 𝑥+2﷯=0 So, 𝑥=2 & 𝑥=−2 Since 𝑥>0 is given, we consider only 𝑥=2 Minimum value of g 𝑥﷯ is g 𝑥﷯= 𝑥﷮2﷯+ 2﷮𝑥﷯ Putting 𝑥=2 g 2﷯= 2﷮2﷯+ 2﷮2﷯ = 1 + 1 = 2 Ex 6.5,3 (Method 2) 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: (vi) g (𝑥) = 𝑥﷮2﷯ + 2﷮𝑥﷯ , 𝑥 > 0 g (𝑥) = 𝑥﷮2﷯ + 2﷮𝑥﷯ , 𝑥 > 0 Finding g’ 𝑥﷯ g’ 𝑥﷯= 𝑑﷮𝑑𝑥﷯ 𝑥﷮2﷯+ 2﷮𝑥﷯﷯ = 𝑑﷮𝑑𝑥﷯ 𝑥﷮2﷯﷯+ 𝑑﷮𝑑𝑥﷯ 2 𝑥﷮−1﷯﷯ = 1﷮2﷯−2 𝑥﷮−2﷯ = 1﷮2﷯− 2﷮ 𝑥﷮2﷯﷯ Putting g’ 𝑥﷯=0 1﷮2﷯− 2﷮ 𝑥﷮2﷯﷯=0 𝑥﷮2﷯− 4﷮2 𝑥﷮2﷯﷯=0 𝑥﷮2﷯−4=0 ×2 𝑥﷮2﷯ 𝑥−2﷯ 𝑥+2﷯=0 So, 𝑥=2 & 𝑥=−2 Since 𝑥>0 is given, we consider only 𝑥=2 Finding g’’ 𝒙﷯ g’ 𝑥﷯= 1﷮2﷯− 2﷮ 𝑥﷮2﷯﷯ g’’ 𝑥﷯= 𝑑﷮𝑑𝑥﷯ 1﷮2﷯− 2﷮ 𝑥﷮2﷯﷯﷯ = 0 – 2 . −2﷯ 𝑥﷮−2−1﷯ = 4 𝑥﷮−3﷯ = 4﷮ 𝑥﷮3﷯﷯ Putting value 𝑥=2 in g’’ 𝑥﷯ g’’ 𝑥﷯= 4﷮ 2﷯﷮3﷯﷯= 4﷮8﷯= 1﷮2﷯ Since g’’ 𝑥﷯>0 when 𝑥=2 ⇒ 𝑥=2 is Point of Local Minima ⇒ g 𝑥﷯ is Minimum at 𝒙=𝟐 Minimum value of g 𝑥﷯ is g 𝑥﷯= 𝑥﷮2﷯+ 2﷮𝑥﷯ Putting 𝑥=2 g 2﷯= 2﷮2﷯+ 2﷮2﷯ = 1 + 1 = 2 Ex 6.5,3 (Method 1) 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: (vii) g (𝑥) = 1﷮ 𝑥﷮2﷯ + 2﷯ Step 1: Finding g’ 𝑥﷯ g’ 𝑥﷯= 𝑑﷮𝑑𝑥﷯ 1﷮ 𝑥﷮2﷯ + 2﷯﷯ g′ 𝑥﷯= 𝑑 1﷯﷮𝑑𝑥﷯ . 𝑥﷮2﷯ + 2﷯ − 𝑑 𝑥﷮2﷯ + 2﷯﷮𝑑𝑥﷯ . 1﷮ 𝑥﷮2﷯ + 2﷯﷮2﷯﷯ g′ 𝑥﷯= 0 . 𝑥﷮2﷯ + 2﷯ − 2𝑥 + 0﷯﷮ 𝑥﷮2﷯ + 2﷯﷮2﷯﷯ g′ 𝑥﷯= 0 − 2𝑥 ﷮ 𝑥﷮2﷯ + 2﷯﷮2﷯﷯ g′ 𝑥﷯= −2𝑥 ﷮ 𝑥﷮2﷯ + 2﷯﷮2﷯﷯ Step 2: Putting g’ 𝑥﷯=0 −2𝑥 ﷮ 𝑥﷮2﷯+2﷯﷮2﷯﷯=0 –2𝑥=0 × 𝑥﷮2﷯+2﷯﷮2﷯ –2𝑥=0 ⇒ 𝑥=0 Step 3: ∴ 𝑥 = 0 is point of local maxima g 𝑥﷯ is maximum at 𝒙 = 0 Maximum value of g 𝑥﷯ g 𝑥﷯= 1﷮ 𝑥﷮2﷯ + 2﷯ Putting 𝑥 = 0 g 0﷯= 1﷮ 0﷮2﷯ + 2﷯ = 1﷮2﷯ Maximum value is 𝟏﷮𝟐﷯ Ex 6.5,3 (Method 2) 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: (vii) g (𝑥) = 1﷮ 𝑥﷮2﷯ + 2﷯ Step 1: Finding g’ 𝑥﷯ g’ 𝑥﷯= 𝑑﷮𝑑𝑥﷯ 1﷮ 𝑥﷮2﷯ + 2﷯﷯ g′ 𝑥﷯= 𝑑 1﷯﷮𝑑𝑥﷯ . 𝑥﷮2﷯ + 2﷯ − 𝑑 𝑥﷮2﷯ + 2﷯﷮𝑑𝑥﷯ . 1﷮ 𝑥﷮2﷯ + 2﷯﷮2﷯﷯ g′ 𝑥﷯= 0 . 𝑥﷮2﷯ + 2﷯ − 2𝑥 + 0﷯﷮ 𝑥﷮2﷯ + 2﷯﷮2﷯﷯ g′ 𝑥﷯= 0 − 2𝑥 ﷮ 𝑥﷮2﷯ + 2﷯﷮2﷯﷯ g′ 𝑥﷯= −2𝑥 ﷮ 𝑥﷮2﷯ + 2﷯﷮2﷯﷯ Step 2: Putting g’ 𝑥﷯=0 −2𝑥 ﷮ 𝑥﷮2﷯+2﷯﷮2﷯﷯=0 –2𝑥=0 × 𝑥﷮2﷯+2﷯﷮2﷯ –2𝑥=0 ⇒ 𝑥=0 Step 3: Finding g’’ 𝑥﷯ g’ 𝑥﷯= −2𝑥﷮ 𝑥﷮2﷯+2﷯﷮2﷯﷯ g’ 𝑥﷯= 𝑑 −2𝑥﷯﷮𝑑𝑥﷯ . 𝑥﷮2﷯+2﷯﷮2﷯ − 𝑑 𝑥﷮2﷯+2﷯﷮2﷯﷮𝑑𝑥﷯ . −2𝑥﷯﷮ 𝑥﷮2﷯+2﷯﷮2﷯﷯﷮2﷯﷯ = −2 𝑥﷮2﷯+2﷯﷮2﷯−2 𝑥﷮2﷯+2﷯﷮2−1﷯. 𝑑 𝑥﷮2﷯+2﷯﷮𝑑𝑥﷯ . −2𝑥﷯﷮ 𝑥﷮2﷯+2﷯﷮2﷯﷯﷮2﷯﷯ = −2 𝑥﷮2﷯+2﷯﷮2﷯−2 𝑥﷮2﷯+2﷯ 2𝑥+0﷯ −2𝑥﷯﷮ 𝑥﷮2﷯+2﷯﷮4﷯﷯ = −2 𝑥﷮2﷯+2﷯﷮2﷯−2 𝑥﷮2﷯+2﷯ 2𝑥﷯ −2𝑥﷯﷮ 𝑥﷮2﷯+2﷯﷮4﷯﷯ = −2 𝑥﷮2﷯+2﷯﷮2﷯+8 𝑥﷮2﷯ 𝑥﷮2﷯+2﷯﷮ 𝑥﷮2﷯+2﷯﷮4﷯﷯ = −2 𝑥﷮2﷯+2﷯ 𝑥﷮2﷯+2﷯−4 𝑥﷮2﷯﷯﷮ 𝑥﷮2﷯+2﷯﷮4﷯﷯ = −2 𝑥﷮2﷯+2﷯ −3 𝑥﷮2﷯+2﷯﷮ 𝑥﷮2﷯+2﷯﷮4﷯﷯ = −2 −3 𝑥﷮2﷯+2﷯﷮ 𝑥﷮2﷯+2﷯﷮3﷯﷯ Putting x = 0 in g’’(x) g’’ 0﷯= −2 −3 0﷯ + 2﷯﷮ 0﷮2﷯+ 2﷯﷮3﷯﷯= −2 0 + 2﷯﷮ 2﷯﷮3﷯﷯= −4﷮8﷯= −1﷮2﷯ Hence g’’ 𝑥﷯<0 when 𝑥 = 0 ⇒ 𝑥 = 0 is point of local maxima ⇒ g 𝑥﷯ is maximum at 𝑥 = 0 Maximum value of g 𝑥﷯ g 𝑥﷯= 1﷮ 𝑥﷮2﷯+2﷯ Putting 𝑥 = 0 g 0﷯= 1﷮ 0﷮2﷯+2﷯ = 1﷮2﷯ Maximum value is 𝟏﷮𝟐﷯ Ex 6.5,3 (Method 1) 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: (viii) f(𝑥) = 𝑥 ﷮1−𝑥﷯, 𝑥 > 0 Step 1: Finding f’ 𝑥﷯ f’ 𝑥﷯= 𝑑 𝑥 ﷮1 − 𝑥﷯﷯﷮𝑑𝑥﷯ f’ 𝑥﷯= 𝑑 𝑥﷯﷮𝑑𝑥﷯ . ﷮1−𝑥﷯ + 𝑑 ﷮1 − 𝑥﷯﷯﷮𝑑𝑥﷯ . 𝑥 = 1 . ﷮1−𝑥﷯ + 1﷮2 ﷮1 − 𝑥﷯﷯ . 𝑑 1 − 𝑥﷯﷮𝑑𝑥﷯ . 𝑥 = ﷮1−𝑥﷯ + 1﷮2 ﷮1 − 𝑥﷯﷯ 0 −1﷯ . 𝑥 = ﷮1−𝑥﷯ – 𝑥﷮2 ﷮1 − 𝑥﷯﷯ = 2 ﷮1 − 𝑥﷯ ﷯﷮2﷯− 𝑥﷮2 ﷮1 − 𝑥﷯﷯ = 2 1 − 𝑥﷯ − 𝑥﷮2 ﷮1 − 𝑥﷯﷯ = 2 − 2𝑥 − 𝑥﷮2 ﷮1 − 𝑥﷯﷯ = 2 − 3𝑥﷮2 ﷮1 − 𝑥﷯﷯ Step 2: Putting f’ 𝑥﷯=0 2 − 3𝑥﷮2 ﷮1 − 𝑥﷯﷯=0 2 – 3𝑥 = 0 × 2 ﷮1−𝑥﷯ 2 – 3𝑥=0 – 3𝑥=−2 𝑥 = 2﷮3﷯ Step 3: ⇒ 𝑥 = 2﷮3﷯ is point of local maxima ⇒ f 𝑥﷯ is maximum at 𝑥 = 2﷮3﷯ Step 4: Finding Maximum value of f 𝑥﷯=𝑥 ﷮1−𝑥﷯ Putting 𝑥= 2﷮3﷯ f 2﷮3﷯﷯= 2﷮3﷯ ﷮1− 2﷮3﷯﷯ = 2﷮3﷯ ﷮ 3−2﷮3﷯﷯ = 2﷮3﷯ ﷮ 1﷮3﷯﷯ = 2﷮3 ﷮3﷯﷯ = 2﷮3 ﷮3﷯﷯ × ﷮3﷯﷮ ﷮3﷯﷯ = 2 ﷮3﷯﷮9﷯` Maximum value of f 𝒙﷯ is 𝟐 ﷮𝟑﷯﷮𝟗﷯ Ex 6.5,3 (Method 2) 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: (viii) f(𝑥) = 𝑥﷐﷮1−𝑥﷯, 𝑥 > 0 Step 1: Finding f’﷐𝑥﷯ f’﷐𝑥﷯=﷐𝑑﷐𝑥﷐﷮1 − 𝑥﷯﷯﷮𝑑𝑥﷯ f’﷐𝑥﷯=﷐𝑑﷐𝑥﷯﷮𝑑𝑥﷯ . ﷐﷮1−𝑥﷯ + ﷐𝑑﷐﷐﷮1 − 𝑥﷯﷯﷮𝑑𝑥﷯ . 𝑥 = 1 . ﷐﷮1−𝑥﷯ + ﷐1﷮2﷐﷮1 − 𝑥﷯﷯ . ﷐𝑑﷐1 − 𝑥﷯﷮𝑑𝑥﷯ . 𝑥 = ﷐﷮1−𝑥﷯ + ﷐1﷮2﷐﷮1 − 𝑥﷯﷯ ﷐0 −1﷯ . 𝑥 = ﷐﷮1−𝑥﷯ – ﷐𝑥﷮2﷐﷮1 − 𝑥﷯﷯ = ﷐2﷐﷐﷐﷮1 − 𝑥﷯ ﷯﷮2﷯− 𝑥﷮2﷐﷮1 − 𝑥﷯﷯ = ﷐2﷐1 − 𝑥﷯ − 𝑥﷮2﷐﷮1 − 𝑥﷯﷯ = ﷐2 − 2𝑥 − 𝑥﷮2﷐﷮1 − 𝑥﷯﷯ = ﷐2 − 3𝑥﷮2﷐﷮1 − 𝑥﷯﷯ Step 2: Putting f’﷐𝑥﷯=0 ﷐2 − 3𝑥﷮2﷐﷮1 − 𝑥﷯﷯=0 2 – 3𝑥 = 0 × 2﷐﷮1−𝑥﷯ 2 – 3𝑥=0 – 3𝑥=−2 𝑥 =﷐2﷮3﷯ Step 3: Finding f’’﷐𝑥﷯ f’﷐𝑥﷯=﷐2−3𝑥﷮2﷐﷮1−𝑥﷯﷯ f’’﷐𝑥﷯=﷐𝑑﷮𝑑𝑥﷯﷐﷐2 − 3𝑥﷮2﷐﷮1 − 𝑥﷯﷯﷯ = ﷐1﷮2﷯﷐﷐﷐𝑑﷐2 − 3𝑥﷯﷮𝑑𝑥﷯ . ﷐﷮1 − 𝑥﷯ − ﷐𝑑﷐﷐﷮1 − 𝑥﷯﷯﷮𝑑𝑥﷯ . ﷐2 − 3𝑥﷯﷮﷐﷐﷐﷮1 − 𝑥﷯﷯﷮2﷯﷯﷯ = ﷐1﷮2﷯﷐﷐﷐0 − 3﷯﷐﷮1 − 𝑥﷯ − ﷐1﷮2﷐﷮1 − 𝑥﷯﷯ . ﷐𝑑﷐1 − 𝑥﷯﷮𝑑𝑥﷯ . ﷐2 − 3𝑥﷯﷮﷐1 − 𝑥﷯﷯﷯ = ﷐1﷮2﷯﷐﷐−3﷐﷮1 − 𝑥﷯ − ﷐1﷮2﷐﷮1 − 𝑥﷯﷯ ﷐0 − 1﷯ . ﷐2 − 3𝑥﷯﷮﷐1 − 𝑥﷯﷯﷯ = ﷐1﷮2﷯﷐﷐−3﷐﷮1 − 𝑥﷯ + ﷐2 − 3𝑥﷮2﷐﷮1 − 𝑥﷯﷯ ﷮1 − 𝑥﷯﷯ = ﷐1﷮2﷯﷐﷐﷐−3﷐﷮1 − 𝑥﷯﷯ ﷐2﷐﷮1 − 𝑥﷯﷯ + 2 − 3𝑥 ﷮2﷐1 − 𝑥﷯﷐﷮1 − 𝑥﷯﷯﷯ = ﷐1﷮2﷯﷐﷐−6﷐1 − 𝑥﷯ + 2 − 3𝑥 ﷮2﷐1 − 𝑥﷯﷐﷮1 − 𝑥﷯﷯﷯ = ﷐1﷮2﷯﷐﷐−6 + 6𝑥 + 2 − 3𝑥 ﷮2﷐1 − 𝑥﷯﷐﷮1 − 𝑥﷯﷯﷯ = ﷐1﷮4﷯﷐﷐−4 + 3𝑥 ﷮﷐﷐1 + 𝑥﷯﷮﷐3﷮2﷯﷯﷯﷯ Hence f’’﷐𝑥﷯=﷐1﷮4﷯﷐﷐−4 + 3𝑥 ﷮﷐﷐1 + 𝑥﷯﷮﷐3﷮2﷯﷯﷯﷯ Putting 𝑥=﷐2﷮3﷯ f’’﷐﷐2﷮3﷯﷯=﷐1﷮4﷯﷐﷐−4 + 3﷐﷐2﷮3﷯﷯﷮﷐﷐1 + ﷐2﷮3﷯﷯﷮﷐3﷮2﷯﷯﷯﷯ =﷐1﷮4﷯﷐﷐−4 + 2﷮﷐﷐﷐5﷮3﷯﷯﷮﷐3﷮2﷯﷯﷯﷯ =﷐1﷮4﷯﷐﷐−2﷮﷐﷐﷐5﷮3﷯﷯﷮﷐3﷮2﷯﷯﷯﷯ = ﷐− 1﷮2﷯﷐﷐﷐3﷮5﷯﷯﷮﷐3﷮2﷯﷯ < 0 ∴ f’’﷐𝑥﷯<0 when 𝑥 = ﷐2﷮3﷯ Hence, 𝑥=﷐2﷮3﷯ is the maxima Step 4: Finding Maximum value of f﷐𝑥﷯=𝑥﷐﷮1−𝑥﷯ Putting 𝑥=﷐2﷮3﷯ f﷐﷐2﷮3﷯﷯=﷐2﷮3﷯﷐﷮1−﷐2﷮3﷯﷯ =﷐2﷮3﷯﷐﷮﷐3 − 2﷮3﷯﷯ =﷐2﷮3﷯﷐﷮﷐1﷮3﷯﷯ =﷐2﷮3﷐﷮3﷯﷯ =﷐2﷮3﷐﷮3﷯﷯ × ﷐﷐﷮3﷯﷮﷐﷮3﷯﷯ =﷐2﷐﷮3﷯﷮9﷯` Maximum value of f﷐𝑥﷯ is ﷐𝟐﷐﷮𝟑﷯﷮𝟗﷯ at x = ﷐𝟏﷮𝟑﷯

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CA Maninder Singh
CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .
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