Ex 6.5

Chapter 6 Class 12 Application of Derivatives
Serial order wise

Get live Maths 1-on-1 Classs - Class 6 to 12

### Transcript

Ex 6.5, 1 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (ii) f (π₯) = 9π₯2+12π₯+2Finding fβ(x) f (π₯)=9π₯2+12π₯+2 Diff. w.r.t π₯ fβ(π₯)=18π₯+12 fβ(π₯)=6(3π₯+2) Putting fβ(π)=π 6(3π₯+2)=0 3π₯+2=0 3π₯=β2 π₯=(β2)/( 3) Hence π₯=(β2)/3 is point of minima of f(π₯) Finding minimum value of f(π₯) at π₯=(β2)/3 f(π₯)=9π₯^2+12π₯+2 Putting π₯=(β2)/3 f(π₯)=9((β2)/3)^2+12((β2)/3)+2=9(4/3)β12(2/3)+2=β2 Thus, Minimum value of f(π)=βπ There is no maximum value Ex 6.5, 1 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (ii) f (π₯) = 9π₯2+12π₯+2Finding fβ(π) f (π₯)=9π₯2+12π₯+2 Diff. w.r.t π₯ fβ(π₯)=π(9π₯^2 + 12π₯ + 2)/ππ₯ fβ(π₯)=18π₯+12 fβ(π₯)=6(3π₯+2) Putting fβ(π)=π 6(3π₯+2)=0 3π₯+2=0 3π₯=β2 π₯=(β2)/3 Finding fββ(π) fβ(π₯)= 6(3π₯+2) Again diff w.r.t π₯ fββ(π₯)=π(6(3π₯+2))/ππ₯ fββ(π₯)=6 π(3π₯ + 2)/ππ₯ fββ(π₯)=6(3+0) fββ(π₯)=6(3) fββ(π₯)=18 So, fββ((β2)/3)=18 Since fββ(π₯)>0 is for π₯=(β2)/3 π₯=(β2)/3 is point of local minima Finding minimum value Putting π₯=(β2)/3 in f(π₯) f (π₯)=9π₯2+12π₯+2 f ((β2)/3)=9((β2)/3)^2+12((β2)/3)+2 =9(4/9)+12((β2)/3)+2 =4β8+2 =β2 Hence, minimum value = β2 There is no maximum value