Examples

Chapter 6 Class 12 Application of Derivatives
Serial order wise

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Example 20 Find local maximum and local minimum values of the function f given by f (π₯)=3π₯4 + 4π₯3 β 12π₯2 + 12f (π₯)=3π₯4 + 4π₯3 β 12π₯2 + 12 Finding fβ (π) fβ (π₯)=π(3π₯4 + 4π₯3 β 12π₯2 + 12)/ππ₯ fβ (π₯)=12π₯^3+12π₯^2 β 24π₯ "+ 0" fβ (π₯)=12(π₯^3+π₯^2β2π₯) fβ (π₯)=12π₯(π₯^2+π₯β2) fβ (π₯)=12π₯ (π₯^2+2π₯βπ₯β2) fβ (π₯)=12π₯ (π₯(π₯+2)β1(π₯+2)) fβ (π₯)=πππ (πβπ)(π+π) Putting fβ (π)=π 12π₯ (π₯β1)(π₯+2)=0 π₯ (π₯β1)(π₯+2)=0 So, π=π,π₯=π,& π₯=βπ Finding fββ(π) f β(π₯)=12(π₯^3+π₯^2β2π₯) f ββ(π₯)=12π(π₯^3 + π₯^2 β 2π₯)/ππ₯ f ββ(π₯)=ππ(ππ^π+ππβπ) At π=π f ββ(0)=12(3(0)^2+2(0)β2)= 32 (0+0 β2)= β 64 < 0 Since fββ(π₯)<0 at π₯=0 β΄ π₯ = 0 is point of local maxima Thus, f(π₯) is maximum at π₯=0 At π=π fββ(1)=12(3(1)^2+2(1)β2)= 12 (3+2β2) = 36 > 0 Since fββ(π₯)>0 at π₯=1 β΄ π₯ = 1 is point of local minima Thus, f(π₯) is minimum at π₯=1 At π=βπ fββ(β2)=12(3(β2)^2+2(β2)β2)= 12 (12β4β2)= 72 > 0 Since fββ(π₯)>0 at π₯=β2 β΄ π₯ = β2 is point of local minima Thus, f(π₯) is minimum at π₯=β2 Finding local minimum and maximum value fβ (π₯)=πππ (πβπ)(π+π) Local maximum value of f (π₯) at π₯=0 f (0)=3(0)4 + 4(0)3 β 12(0)2 + 12 = 0 + 0 β 0 + 12 = 12 Local minimum value of f (π₯) at π₯=1 f (1)=3(1)4 + 4(1)3 β 12(1)2 + 12 = 3 + 4 β 12 + 12 = 7 Local Minimum value of f (π₯) at π₯=β2 f (β2)=3(β2)4 + 4(β2)3 β 12(β2)2 + 12 = 48 β 32 β 48 + 12 = β 20