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Example 39 - Find absolute maximum, minimum values of f(x)

Example 39 - Chapter 6 Class 12 Application of Derivatives - Part 2
Example 39 - Chapter 6 Class 12 Application of Derivatives - Part 3

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Example 39 Find the absolute maximum and minimum values of a function f given by 𝑓 (π‘₯) = 2π‘₯3 – 15π‘₯2 + 36π‘₯ +1 on the interval [1, 5]. f(π‘₯) = 2π‘₯3 – 15π‘₯2 + 36π‘₯ + 1 Finding f’(𝒙) f’(π‘₯)=6π‘₯^2βˆ’30π‘₯+36 Putting f ’(𝒙)=𝟎 6π‘₯2 – 30π‘₯ + 36 = 0 6(π‘₯^2βˆ’5π‘₯+6)=0 (π‘₯^2βˆ’5π‘₯+6)=0 (π‘₯^2βˆ’2π‘₯βˆ’3π‘₯+6)=0 π‘₯(π‘₯βˆ’2)βˆ’3(π‘₯βˆ’2)=0 (π‘₯βˆ’2)(π‘₯βˆ’3)=0 Hence. 𝒙 = 2 or 𝒙 = 3 Since we are given interval [𝟏 , πŸ“] Hence, calculating f(π‘₯) at π‘₯ = 1, 2, 3, 5 Hence, Absolute maximum value is 56 at 𝒙=πŸ“ Absolute minimum value is 24 at 𝒙=𝟏

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