Slide35.JPG

Slide36.JPG
Slide37.JPG

This video is only available for Teachoo black users

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Example 27 Find the absolute maximum and minimum values of a function f given by 𝑓 (𝑥) = 2𝑥3 – 15𝑥2 + 36𝑥 +1 on the interval [1, 5]. f(𝑥) = 2𝑥3 – 15𝑥2 + 36𝑥 + 1 Finding f’(𝒙) f’(𝑥)=6𝑥^2−30𝑥+36 Putting f ’(𝒙)=𝟎 6𝑥2 – 30𝑥 + 36 = 0 6(𝑥^2−5𝑥+6)=0 (𝑥^2−5𝑥+6)=0 (𝑥^2−2𝑥−3𝑥+6)=0 𝑥(𝑥−2)−3(𝑥−2)=0 (𝑥−2)(𝑥−3)=0 Hence. 𝒙 = 2 or 𝒙 = 3 Since we are given interval [𝟏 , 𝟓] Hence, calculating f(𝑥) at 𝑥 = 1, 2, 3, 5 Hence, Absolute maximum value is 56 at 𝒙=𝟓 Absolute minimum value is 24 at 𝒙=𝟏

Ask a doubt
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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.