Slide26.JPG

Slide27.JPG
Slide28.JPG Slide29.JPG Slide30.JPG Slide31.JPG Slide32.JPG

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

Transcript

Ex 6.5, 2 Find the maximum and minimum values, if any, of the following functions given by (i) 𝑓 (π‘₯)=|π‘₯ + 2| –1 𝑓 (π‘₯)=|π‘₯ + 2| –1 Minimum value of |π‘₯ + 2|=0 Minimum value of f(π‘₯)= minimum value of |π‘₯ + 2| –1 =0βˆ’1 =βˆ’1 Hence minimum value of f(𝒙)=βˆ’πŸ And there is no maximum value of f(x) Ex 6.5, 2 Find the maximum and minimum values, if any, of the following functions given by (ii) 𝑓(π‘₯)= – | π‘₯ +1|+3 f(π‘₯)= – | π‘₯ +1|+3 We know that | π‘₯ +1|β‰₯0 So, βˆ’| π‘₯ +1|≀0 Maximum value of g(π‘₯) = maximum value of – | π‘₯ +1|+3 = 0 + 3 = 3 Hence maximum value of f(x) is 3 And there is no minimum value of f(x) Ex 6.5, 2 Find the maximum and minimum values, if any, of the following functions given by (iii) β„Ž(π‘₯)= sin ⁑(2π‘₯)+ 5 β„Ž(π‘₯)= sin ⁑(2π‘₯)+ 5 We know that –1 ≀ sin ΞΈ ≀ 1 –1 ≀ sin 2π‘₯ ≀ 1 Adding 5 both sides –1 + 5 ≀ sin2π‘₯ + 5 ≀ 1 + 5 4 ≀ sin 2π‘₯ + 5 ≀ 6 4 ≀ f(π‘₯)≀6 Hence Maximum value of f(𝒙)=πŸ” & Minimum value of f(𝒙)=πŸ’ Ex 6.5, 2 Find the maximum and minimum values, if any, of the following functions given by (iv) 𝑓 (π‘₯)=|sin⁑4π‘₯+3| 𝑓 (π‘₯)=| sin⁑4π‘₯+3| We know that –1 ≀ sin ΞΈ ≀ 1 So, –1 ≀ sin 4π‘₯ ≀ 1 Adding 3 both sides –1 + 3 ≀ sin 4π‘₯ + 3 ≀ 1 + 3 2 ≀ sin 4π‘₯ +3 ≀ 4 Taking modulus |2| ≀ | sin⁑4π‘₯+3| ≀ |4| 2 ≀ | sin⁑4π‘₯+3| ≀ |4| 2 ≀ f(π‘₯)≀4 Hence Maximum value of f(𝒙) is 4 & Minimum value of f(𝒙) is 2 Ex 6.5, 2 Find the maximum and minimum values, if any, of the following functions given by (v) β„Ž(π‘₯)=π‘₯ + 1 , π‘₯ ∈ (βˆ’1 , 1) Drawing graph of f(π‘₯)=π‘₯+1 β„Ž(π‘₯) have Maximum value of point closest to x = 1 & Minimum value of point closest to x = –1 but its not possible to locate such points Thus the given function has neither the maximum value nor minimum value

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 has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.