# Ex 6.5, 1 (iii) - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at Aug. 19, 2021 by Teachoo

Ex 6.5

Ex 6.5, 1 (i)
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Ex 6.5, 1 (iii) Important You are here

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Last updated at Aug. 19, 2021 by Teachoo

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Ex 6.5,1 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (iii) π (π₯) = β(π₯ β 1)^2+10 f(π₯)=β(π₯β1)^2+10 Finding fβ (x) Diff w.r.t π₯ fβ(π₯)=π(β(π₯β1)^2+10)/ππ₯ fβ(π₯) = β2(π₯β1)(π(π₯β1)/ππ₯)+0 fβ(π₯) = β2(π₯β1)(1β0) + 0 fβ(π₯)=β2(π₯β1) Putting fβ(π)=π β2(π₯β1)=0 (π₯β1)=0 π₯=1 Hence, π₯=1 is point of Maxima & No point of Minima Thus, f(π₯) has maximum value at π₯=1 Putting x=1 in f(x) f(π₯)=β(π₯β1)^2+10 f(1)=β(1β1)^2+10 = 0 + 10 = 10 Maximum value of f(π₯) is 10 There is no minimum value of f(π) Ex 6.5,1 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (iii) π (π₯) = β (π₯ β 1)^2+10 f(π₯)=β(π₯β1)^2+10 Finding fβ(x) Diff w.r.t π₯ fβ(π₯)=π(β(π₯β1)^2+10)/ππ₯ fβ(π₯) = β2(π₯β1)(π(π₯β1)/ππ₯)+0 fβ(π₯) = β2(π₯β1)(1β0) + 0 fβ(π₯)=β2(π₯β1) Putting fβ(π)=π β2(π₯β1)=0 (π₯β1)=0 π₯=1 Finding fββ(π) fβ(π₯)=β2(π₯β1) Again diff w.r.t π₯ fββ(π₯)=π(β2(π₯ β 1))/ππ₯ =β2 π(π₯ β 1)/ππ₯ =β2(1β0) =β2 Since fββ(π₯) < 0 for π₯=1 Hence, f(π₯) has Maximum value at π₯=1 Finding maximum value of f(π) f(π₯)=β(π₯β1)^2+10 Putting π₯=1 f(π₯) =β(1β1)^2+10 = 0 + 10 = 10 Maximum value of f(π) is 10 There is no minimum value of f(π)