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


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 - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.