# Example 47

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Example 47 Find intervals in which the function given by f(𝑥) =310𝑥4 – 45𝑥3– 3𝑥2 + 365𝑥 + 11 is (a) strictly increasing (b) strictly decreasing f𝑥 = 310𝑥4 – 45𝑥3– 3𝑥2 + 365𝑥 + 11 Step 1: Finding f’𝑥 f’𝑥 = 310 × 4𝑥3 – 45 × 3𝑥2 – 3 × 2x + 365 + 0 f’𝑥 = 1210𝑥3– 125𝑥2– 6x + 365 f’𝑥 = 65𝑥3− 125𝑥2– 6x + 365 f’𝑥 = 6𝑥35−2𝑥25−𝑥+65 f’𝑥 = 6𝑥3 − 2𝑥2− 5𝑥 + 65 = 65 𝑥3−2𝑥2−5𝑥+6 = 65𝑥−1𝑥2−𝑥−6 = 65 𝑥−1𝑥2−3𝑥+2𝑥−6 = 65 𝑥−1𝑥𝑥−3+2𝑥−3 = 65 𝑥−1𝑥+2𝑥−3 Hence f’𝑥 = 65𝑥−1𝑥+2𝑥−3 Step 2: Putting f’𝑥=0 65𝑥−1𝑥+2𝑥−3 = 0 𝑥−1𝑥+2𝑥−3 = 0 Hence x = –2 , 1 & 3 Step 3: Plotting point on real line Thus, we get four disjoint intervals i.e. −−2 ,−2, 1 ,1 , 3, 3 , uc1 ⇒ f𝑥 is strictly decreasing on the interval 𝑥 ∈−−𝟐& 𝟏 , 𝟑 f𝑥 is strictly increasing on the interval 𝑥 ∈−𝟐,𝟏 & 𝟑 , uc1

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Chapter 6 Class 12 Application of Derivatives

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About the Author

CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .