Check sibling questions

Ex 6.5, 10 - Find max value of 2x3 - 24 x + 107 in [1, 3]

Ex 6.5,10 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.5,10 - Chapter 6 Class 12 Application of Derivatives - Part 3
Ex 6.5,10 - Chapter 6 Class 12 Application of Derivatives - Part 4

Now learn Economics at Teachoo for Class 12


Transcript

Ex 6.5, 10 Find the maximum value of 2π‘₯3 – 24π‘₯ + 107 in the interval [1, 3]. Find the maximum value of the same function in [–3, –1]. Let f(π‘₯)=2π‘₯^3βˆ’24π‘₯+107 Finding f’(π‘₯) f’(π‘₯)=𝑑(2π‘₯^3 βˆ’ 24π‘₯ + 107)/𝑑π‘₯ = 2 Γ— 3π‘₯^2βˆ’24 = 6π‘₯^2βˆ’24 = 6 (π‘₯^2βˆ’4" " ) Putting f(π‘₯)=0 6 (π‘₯^2βˆ’4" " )=0 π‘₯^2βˆ’4" = 0 " π‘₯^2=4 π‘₯=±√4 π‘₯=Β±2 Thus, π‘₯=2 , – 2 Since x is in interval [1 , 3] π‘₯ = 2 is only Critical point Also, since given the interval π‘₯= ∈ [1 , 3] We calculate f(x) at π‘₯= 1 , 2 & 3 Hence maximum value of f(𝒙)=πŸ–πŸ— at 𝒙 = 3 in the interval [1 , 3] For the interval [βˆ’πŸ‘ , βˆ’πŸ] π‘₯ = –2 is only Critical Point Also, since given the interval π‘₯= ∈ [βˆ’3,βˆ’1] We calculate f(x) at π‘₯= –1 , –2 & –3 Hence maximum value of f(𝒙)=πŸπŸ‘πŸ— at 𝒙 = –2 in the interval [βˆ’3,βˆ’1]

Davneet Singh's photo - Teacher, Engineer, Marketer

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.