Example 11 - Class 12
Last updated at Dec. 8, 2016 by Teachoo
Transcript
Example 11 Find the intervals in which the function f given by f 𝑥=4𝑥3−6𝑥2–72𝑥+30 is (a) strictly increasing (b) strictly decreasing. f 𝑥=4𝑥3−6𝑥2 –72𝑥+30 Step 1: Calculating f’(x) f 𝑥=4𝑥3−6𝑥2–72𝑥+30 𝑓′𝑥=12𝑥2−12𝑥 –72𝑥 𝑓′𝑥 =12𝑥2−𝑥 –6 𝑓′𝑥 =12𝑥2−2𝑥 −3𝑥 –6 𝑓′(𝑥) =12(𝑥(𝑥 − 2) − 3(𝑥 − 2)) 𝑓′𝑥=12𝑥−2𝑥 –3 Step 2: Putting f’(x) = 0 12𝑥−2𝑥−3=0 𝑥−2𝑥−3=0 So, x = 2 and x = 3 Step 3: Plotting point on real line The points divide the real line into 3 disjoint intervals. i.e. − 2 2 ,3 & 3 , uc1 Step 4: (a) f is strictly increasing in − ,𝟐 & 𝟑 ,uc1 (b) f is strictly decreasing in 𝟐, 𝟑
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Example 11 You are here
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Chapter 6 Class 12 Application of Derivatives
Serial order wise
About the Author
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.
