# Ex 6.5,1

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 6.5,1 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (i) f (𝑥) = 2𝑥 – 12 + 3 f 𝑥= 2𝑥−12+3 Hence, Minimum value of 2𝑥−12 = 0 Minimum value of 2𝑥− 12+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 – 12+3 Step 1: Finding f’(x) f 𝑥= 2𝑥−12+3 f’ 𝑥= 2 2𝑥−1 Step 2: Putting f’ 𝑥=0 2 2𝑥−1=0 2𝑥 – 1 = 0 2𝑥 = 1 𝑥 = 12 Step 3: Thus, x = 12 is the minima Finding minimum value f 𝑥= 2𝑥−12+3 Putting 𝑥 = 12 f 12= 2 × 12−12+3= 1−12+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 – 12 + 3 Double Derivative Test f 𝑥= 2𝑥−12+3 f’ 𝑥=2 2𝑥−12−1 = 2 2𝑥−1 Putting f’ 𝑥=0 2 2𝑥−1=0 2𝑥−1=0 2𝑥 = 0 + 1 𝑥 = 12 Now, finding f’’ 𝑥 f’ 𝑥=2 2𝑥−1 f’ 𝑥 = 4𝑥 – 2 f’’ 𝑥= 4 So, f’’ 12 = 4 Since f’’ 12 > 0 , 𝑥 = 12 is point of local minima Putting 𝑥 = 12 , we can calculate minimum value f 𝑥 = 2𝑥−12+3 f 12= 2 12−12+3= 1−12+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= −23 is point of minima of f 𝑥 Finding minimum value of f 𝑥 at 𝑥= −23 f 𝑥=9 𝑥2+12𝑥+2 Putting 𝑥= −23 f 𝑥=9 −232+12 −23+2=9 43−12 23+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 𝑥2 + 12𝑥 + 2𝑑𝑥 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= −23 Step 3: Finding f’’ 𝑥 f’ 𝑥= 6 3𝑥+2 Again diff w.r.t 𝑥 f’’ 𝑥= 𝑑 6 3𝑥+2𝑑𝑥 f’’ 𝑥=6 𝑑 3𝑥 + 2𝑑𝑥 f’’ 𝑥=6 3+0 f’’ 𝑥=6 3 f’’ 𝑥=18 So, f’’ −23=18 Since f’’ 𝑥>0 is for 𝑥= −23 𝑥= −23 is point of local minima Step 4: Putting 𝑥= −23 we can calculate minimum value of f 𝑥 f (𝑥)=9𝑥2+12𝑥+2 f −23=9 −232+12 −23+2 =9 49+12 −23+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 – 12+10 f 𝑥=− 𝑥−12+10 Step 1: Diff w.r.t 𝑥 f’ 𝑥= 𝑑 − 𝑥−12+10𝑑𝑥 f’ 𝑥 = –2 𝑥−1 𝑑 𝑥−1𝑑𝑥+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−12+10 Putting 𝑥=1 f 1=− 1−12+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 – 12+10 f 𝑥=− 𝑥−12+10 Step 1: Diff w.r.t 𝑥 f’ 𝑥= 𝑑 − 𝑥−12+10𝑑𝑥 f’ 𝑥 = –2 𝑥−1 𝑑 𝑥−1𝑑𝑥+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 − 1𝑑𝑥 =−2 𝑑 𝑥 − 1𝑑𝑥 =−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−12+10 Putting 𝑥=1 f 𝑥 =− 1−12+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= 𝑥3+1 Step 1: Finding f’(x) f’ 𝑥= 𝑑 𝑥3+1𝑑𝑥 =3 𝑥2 Step 2: Putting f’ 𝑥=0 3 𝑥2=0 𝑥2=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= 𝑥3+1 Step 1: Finding f’(x) f’ 𝑥= 𝑑 𝑥3+1𝑑𝑥 =3 𝑥2 Step 2: Putting f’ 𝑥=0 3 𝑥2=0 𝑥2=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

Chapter 6 Class 12 Application of Derivatives

Ex 6.3,5
Important

Ex 6.3,7 Important

Ex 6.3,12 Important

Ex 6.3,15 Important

Ex 6.3,26 Important

Example 35 Important

Example 37 Important

Example 38 Important

Example 40 Important

Ex 6.5,1 Important You are here

Ex 6.5,5 Important

Ex 6.5,7 Important

Ex 6.5,11 Important

Ex 6.5,18 Important

Ex 6.5,20 Important

Ex 6.5,23 Important

Ex 6.5,26 Important

Ex 6.5,28 Important

Example 46 Important

Example 47 Important

Misc 6 Important

Misc 11 Important

Misc 13 Important

Misc 17 Important

Misc 22 Important

Class 12

Important Question for exams Class 12

- Chapter 1 Class 12 Relation and Functions
- Chapter 2 Class 12 Inverse Trigonometric Functions
- Chapter 3 Class 12 Matrices
- Chapter 4 Class 12 Determinants
- Chapter 5 Class 12 Continuity and Differentiability
- Chapter 6 Class 12 Application of Derivatives
- Chapter 7 Class 12 Integrals
- Chapter 8 Class 12 Application of Integrals
- Chapter 9 Class 12 Differential Equations
- Chapter 10 Class 12 Vector Algebra
- Chapter 11 Class 12 Three Dimensional Geometry
- Chapter 12 Class 12 Linear Programming
- Chapter 13 Class 12 Probability

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.