Ex 6.2, 5 - Chapter 6 Class 12 Application of Derivatives

Last updated at April 16, 2024 by

Ex 6.2, 5 - Find intervals where f(x) = 2x^3 - 3x^2 - 36x + 7 is

Ex 6.2, 5 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.2, 5 - Chapter 6 Class 12 Application of Derivatives - Part 3

Transcript

Ex 6.2, 5 Find the intervals in which the function f given by f (𝑥) = 2𝑥3 – 3𝑥2 – 36𝑥 + 7 is (a) strictly increasing (b) strictly decreasingf(𝑥) = 2𝑥3 – 3𝑥2 – 36𝑥 + 7 Calculating f’(𝒙) f’(𝑥) = 6𝑥2 – 6𝑥 – 36 + 0 f’(𝑥) = 6 (𝑥2 – 𝑥 – 6 ) f’(𝑥) = 6(𝑥^2 – 3𝑥 + 2𝑥 – 6) f’(𝑥) = 6(𝑥(𝑥 − 3) + 2 (𝑥 − 3)) f’(𝒙) = 6(𝒙 – 3) (𝒙 + 2) Putting f’(x) = 0 6(𝑥+2)(𝑥 –3)=0 (𝑥+2)(𝑥 –3)=0 So, x = −2 and x = 3 Plotting points on number line Hence, f is strictly increasing in (−∞ ,−𝟐) & (𝟑 ,∞) f is strictly decreasing in (−𝟐, 𝟑)

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