Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives

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Ex 6.5,1 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (i) f (𝑥) = 2𝑥 – 1﷯﷮2﷯ + 3 f 𝑥﷯= 2𝑥−1﷯﷮2﷯+3 Hence, Minimum value of 2𝑥−1﷯﷮2﷯ = 0 Minimum value of 2𝑥− 1﷮2﷯﷯+3 = 0 + 3 = 3 Also, there is no maximum value of 𝑥 ∴ There is no maximum value of f(x) Ex 6.5,1 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (i) f (𝑥) = 2𝑥 – 1﷯﷮2﷯+3 Step 1: Finding f’(x) f 𝑥﷯= 2𝑥−1﷯﷮2﷯+3 f’ 𝑥﷯= 2 2𝑥−1﷯ Step 2: Putting f’ 𝑥﷯=0 2 2𝑥−1﷯=0 2𝑥 – 1 = 0 2𝑥 = 1 𝑥 = 1﷮2﷯ Step 3: Thus, x = 1﷮2﷯ is the minima Finding minimum value f 𝑥﷯= 2𝑥−1﷯﷮2﷯+3 Putting 𝑥 = 1﷮2﷯ f 1﷮2﷯﷯= 2 × 1﷮2﷯−1﷯﷮2﷯+3= 1−1﷯﷮2﷯+3= 3 ∴ Minimum value = 3 There is no maximum value Ex 6.5,1 (Method 3) Find the maximum and minimum values, if any, of the following functions given by (i) 𝑓 𝑥﷯= 2𝑥 – 1﷯﷮2﷯ + 3 Double Derivative Test f 𝑥﷯= 2𝑥−1﷯﷮2﷯+3 f’ 𝑥﷯=2 2𝑥−1﷯﷮2−1﷯ = 2 2𝑥−1﷯ Putting f’ 𝑥﷯=0 2 2𝑥−1﷯=0 2𝑥−1﷯=0 2𝑥 = 0 + 1 𝑥 = 1﷮2﷯ Now, finding f’’ 𝑥﷯ f’ 𝑥﷯=2 2𝑥−1﷯ f’ 𝑥﷯ = 4𝑥 – 2 f’’ 𝑥﷯= 4 So, f’’ 1﷮2﷯﷯ = 4 Since f’’ 1﷮2﷯﷯ > 0 , 𝑥 = 1﷮2﷯ is point of local minima Putting 𝑥 = 1﷮2﷯ , we can calculate minimum value f 𝑥﷯ = 2𝑥−1﷯﷮2﷯+3 f 1﷮2﷯﷯= 2 1﷮2﷯﷯−1﷯﷮2﷯+3= 1−1﷯﷮2﷯+3= 3 Hence, Minimum value = 3 There is no Maximum value Ex 6.5,1 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (ii) f (𝑥) = 9𝑥2+12𝑥+2 Step 1: Finding f’(x) f (𝑥)=9𝑥2+12𝑥+2 Diff. w.r.t 𝑥 f’ 𝑥﷯=18𝑥+12 f’ 𝑥﷯=6 3𝑥+2﷯ Step 2: Putting f’ 𝑥﷯=0 6 3𝑥+2﷯=0 3𝑥+2=0 3𝑥=−2 𝑥= −2﷮ 3﷯ Step 3: Hence 𝑥= −2﷮3﷯ is point of minima of f 𝑥﷯ Finding minimum value of f 𝑥﷯ at 𝑥= −2﷮3﷯ f 𝑥﷯=9 𝑥﷮2﷯+12𝑥+2 Putting 𝑥= −2﷮3﷯ f 𝑥﷯=9 −2﷮3﷯﷯﷮2﷯+12 −2﷮3﷯﷯+2=9 4﷮3﷯﷯−12 2﷮3﷯﷯+2=−2 Thus, Minimum value of f 𝒙﷯=−𝟐 There is no maximum value Ex 6.5,1 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (ii) f (𝑥) = 9𝑥2+12𝑥+2 Step 1: f (𝑥)=9𝑥2+12𝑥+2 Diff. w.r.t 𝑥 f’ 𝑥﷯= 𝑑 9 𝑥﷮2﷯ + 12𝑥 + 2﷯﷮𝑑𝑥﷯ f’ 𝑥﷯=18𝑥+12 f’ 𝑥﷯=6 3𝑥+2﷯ Step 2: Putting f’ 𝑥﷯=0 6 3𝑥+2﷯=0 3𝑥+2=0 3𝑥=−2 𝑥= −2﷮3﷯ Step 3: Finding f’’ 𝑥﷯ f’ 𝑥﷯= 6 3𝑥+2﷯ Again diff w.r.t 𝑥 f’’ 𝑥﷯= 𝑑 6 3𝑥+2﷯﷯﷮𝑑𝑥﷯ f’’ 𝑥﷯=6 𝑑 3𝑥 + 2﷯﷮𝑑𝑥﷯ f’’ 𝑥﷯=6 3+0﷯ f’’ 𝑥﷯=6 3﷯ f’’ 𝑥﷯=18 So, f’’ −2﷮3﷯﷯=18 Since f’’ 𝑥﷯>0 is for 𝑥= −2﷮3﷯ 𝑥= −2﷮3﷯ is point of local minima Step 4: Putting 𝑥= −2﷮3﷯ we can calculate minimum value of f 𝑥﷯ f (𝑥)=9𝑥2+12𝑥+2 f −2﷮3﷯﷯=9 −2﷮3﷯﷯﷮2﷯+12 −2﷮3﷯﷯+2 =9 4﷮9﷯﷯+12 −2﷮3﷯﷯+2 =4−8+2 =−2 Hence minimum value = –2 There is no maximum value Ex 6.5,1 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (iii) 𝑓 (𝑥) = – 𝑥 – 1﷯﷮2﷯+10 f 𝑥﷯=− 𝑥−1﷯﷮2﷯+10 Step 1: Diff w.r.t 𝑥 f’ 𝑥﷯= 𝑑 − 𝑥−1﷯﷮2﷯+10﷯﷮𝑑𝑥﷯ f’ 𝑥﷯ = –2 𝑥−1﷯ 𝑑 𝑥−1﷯﷮𝑑𝑥﷯﷯+0 f’ 𝑥﷯ = –2 𝑥−1﷯ 1−0﷯ + 0 f’ 𝑥﷯=−2 𝑥−1﷯ Step 2: Putting f’ 𝑥﷯=0 –2 𝑥−1﷯=0 𝑥−1﷯=0 𝑥=1 Step 3: Hence, 𝑥=1 is point of Maxima & No point of Minima Thus, f 𝑥﷯ has maximum value at 𝑥=1 f 𝑥﷯=− 𝑥−1﷯﷮2﷯+10 Putting 𝑥=1 f 1﷯=− 1−1﷯﷮2﷯+10 = 0 + 10 = 10 Maximum value of f 𝑥﷯ is 10 There is no minimum value of f 𝒙﷯ Ex 6.5,1 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (iii) 𝑓 (𝑥) = – 𝑥 – 1﷯﷮2﷯+10 f 𝑥﷯=− 𝑥−1﷯﷮2﷯+10 Step 1: Diff w.r.t 𝑥 f’ 𝑥﷯= 𝑑 − 𝑥−1﷯﷮2﷯+10﷯﷮𝑑𝑥﷯ f’ 𝑥﷯ = –2 𝑥−1﷯ 𝑑 𝑥−1﷯﷮𝑑𝑥﷯﷯+0 f’ 𝑥﷯ = –2 𝑥−1﷯ 1−0﷯ + 0 f’ 𝑥﷯=−2 𝑥−1﷯ Step 2: Putting f’ 𝑥﷯=0 –2 𝑥−1﷯=0 𝑥−1﷯=0 𝑥=1 Step 3: Finding f’’ 𝑥﷯ f’ 𝑥﷯=−2 𝑥−1﷯ Again diff w.r.t 𝑥 f’’ 𝑥﷯= 𝑑 −2 𝑥 − 1﷯﷯﷮𝑑𝑥﷯ =−2 𝑑 𝑥 − 1﷯﷮𝑑𝑥﷯ =−2 1−0﷯ =−2 Since f’’ 𝑥﷯ < 0 for 𝑥=1 Hence f(𝑥) has Maximum value at 𝑥=1 Finding maximum value of f 𝑥﷯ f 𝑥﷯=− 𝑥−1﷯﷮2﷯+10 Putting 𝑥=1 f 𝑥﷯ =− 1−1﷯﷮2﷯+10 = 0 + 10 = 10 Maximum value of f 𝒙﷯ is 10 There is no minimum value of f 𝒙﷯ Ex 6.5,1 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (iv) f(𝑥) = 𝑥3 + 1 f 𝑥﷯= 𝑥﷮3﷯+1 Step 1: Finding f’(x) f’ 𝑥﷯= 𝑑 𝑥﷮3﷯+1﷯﷮𝑑𝑥﷯ =3 𝑥﷮2﷯ Step 2: Putting f’ 𝑥﷯=0 3 𝑥﷮2﷯=0 𝑥﷮2﷯=0 𝑥=0 Step 3: ⇒ Therefore by first derivate test, the point 𝑥=0 is Neither a point of local maxima nor a point of local Minima Hence 𝒙=𝟎 is point of inflexion Ex 6.5,1 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (iv) f(𝑥) = 𝑥3 + 1 f 𝑥﷯= 𝑥﷮3﷯+1 Step 1: Finding f’(x) f’ 𝑥﷯= 𝑑 𝑥﷮3﷯+1﷯﷮𝑑𝑥﷯ =3 𝑥﷮2﷯ Step 2: Putting f’ 𝑥﷯=0 3 𝑥﷮2﷯=0 𝑥﷮2﷯=0 𝑥=0 Step 3: Finding f’’(x) f’(x) = 3x2 f’’(x) = 6x Finding f’’(x) at x = 0 f’’(0) = 6 × 0 = 0 Since f’’(x) = 0 at x = 0 ⇒ the point 𝑥=0 is Neither a point of local maxima nor a point of local Minima Hence 𝒙=𝟎 is point of inflexion