Ex 6.3

Chapter 6 Class 12 Application of Derivatives
Serial order wise

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### Transcript

Ex 6.3, 7 Find both the maximum value and the minimum value of 3π₯4 β 8π₯3 + 12π₯2 β 48π₯ + 25 on the interval [0, 3].Let f(x) = 3π₯4 β 8π₯3 + 12π₯2 β 48π₯ + 25, where π₯ β [0, 3] Finding fβ(π) fβ(π₯)=π(3π₯^4 β 8π₯^3 + 12π₯^2 β 48π₯ + 25)/ππ₯ fβ(π₯)=3 Γ4π₯^3β8 Γ3π₯^2+12 Γ2π₯β48+0 fβ(π₯)=12π₯^3β24π₯^2+24π₯β48 fβ(π₯)=12(π₯^3β2π₯^2+2π₯β4) Putting fβ(π)=π 12(π₯^3β2π₯^2+2π₯β4)=0 π₯^3β2π₯^2+2π₯β4=0 π₯^2 (π₯β2)+2(π₯β2)=0 (π₯^2+2)(π₯β2)=0 Since π₯^2=β2 is not possible Thus π₯=2 is only critical point Square of number can not be negative π₯^2+2=0 π₯^2=β2 π₯β2=0 π₯=2 Since are given interval π₯ β [0 , 3] Hence , calculating f(π₯) at π₯ = 0 , 2 & 3 Hence, Minimum value of f(π₯) is β39 at π = 2 Maximum value of f(π₯) is 25 at π = 0