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Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 17

Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 18
Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 19

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Ex 6.5, 3 Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be: (v) 𝑓 (π‘₯)=π‘₯3 –6π‘₯2+9π‘₯+15𝑓 (π‘₯)=π‘₯3 –6π‘₯2+9π‘₯+15 Finding f’(𝒙) f’(π‘₯)=𝑑(π‘₯3 – 6π‘₯2 + 9π‘₯ + 15" " )/𝑑π‘₯ f’(π‘₯)=3π‘₯^2βˆ’12π‘₯+9 f’(π‘₯)=3(π‘₯^2βˆ’4π‘₯+3) Putting f’(𝒙)=𝟎 3(π‘₯^2βˆ’4π‘₯+3)=0 π‘₯^2βˆ’4π‘₯+3=0 π‘₯^2βˆ’3π‘₯βˆ’π‘₯+3=0 π‘₯(π‘₯βˆ’3)βˆ’1(π‘₯βˆ’3)=0 (π‘₯βˆ’1)(π‘₯βˆ’3)=0 So, x = 1 & x = 3 Finding f’’(𝒙) f’(π‘₯)=3(π‘₯^2βˆ’4π‘₯+3) f’’(π‘₯)=𝑑(3(π‘₯^2 βˆ’ 4π‘₯+3))/𝑑π‘₯ = 3(2π‘₯βˆ’4+0) = 6π‘₯βˆ’12 Putting 𝒙=𝟏 in f’’(𝒙) f’’(1)=6(1)βˆ’12 = 6 – 12 = – 6 < 0 Since f’’(π‘₯)<0 when π‘₯=1 β‡’ π‘₯=1 is point of local maxima ∴ f(π‘₯) is maximum at 𝒙=𝟏 Maximum value of f(π‘₯) at π‘₯ = 1 f(π‘₯)=π‘₯^3βˆ’6π‘₯^2+9π‘₯+15 f(1)=(1)^3βˆ’6(1)^2+9(1)+15 = 1 – 6 + 9 + 15 = 19 Putting 𝒙=πŸ‘ in f’’(x) f’’(π‘₯)=6π‘₯βˆ’12 f’’(3)=6(3)βˆ’12 = 18 – 12 = 6 > 0 Since f’’(π‘₯)>0 when π‘₯=3 β‡’ π‘₯=3 is point of local minima ∴ f(π‘₯) is minimum at 𝒙=πŸ‘ Minimum value of f(π‘₯) at π‘₯ = 3 f(π‘₯)=π‘₯^3βˆ’6π‘₯^2+9π‘₯+15 f(3)=(3)^3βˆ’6(3)^2+9(3)+15 = 27 – 54 + 27 + 15 = 15

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