web analytics

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

Slide36.JPG
Slide37.JPG Slide38.JPG Slide39.JPG Slide40.JPG Slide41.JPG Slide42.JPG Slide1.JPG Slide2.JPG Slide3.JPG Slide4.JPG Slide5.JPG Slide6.JPG Slide7.JPG Slide8.JPG Slide9.JPG Slide10.JPG Slide11.JPG Slide12.JPG Slide13.JPG Slide14.JPG Slide15.JPG Slide16.JPG Slide17.JPG Slide18.JPG Slide19.JPG Slide20.JPG Slide21.JPG Slide22.JPG Slide23.JPG Slide24.JPG Slide25.JPG Slide26.JPG Slide27.JPG Slide28.JPG Slide29.JPG Slide30.JPG Slide31.JPG Slide32.JPG Slide33.JPG Slide34.JPG Slide35.JPG Slide36.JPG Slide37.JPG Slide38.JPG Slide1.JPG Slide2.JPG Slide3.JPG Slide4.JPG Slide5.JPG Slide6.JPG





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

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﷯

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
Jail