1. Chapter 6 Class 12 Application of Derivatives
2. Serial order wise

Transcript

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﷯