DSE 2013 Q 23

[SoniaKapoor]

[ViV]

using extreme value theorem. Find F'(x) = 0.  you ll get x=3

[Dreyfus]

Since the fun is defined in closed and bounded interval....therefore the extreme value of function can occur either on extreme points ie at -1 or 5 or it can occur at fund's stationary value ie at 3....so evaluate f(-1), f(5) (given) and f(3)....

[SoniaKapoor]

yr after diff i got y=3/x^1/3-x^2/3
wont we consider the case x=0??

[ViV]

in differentiation 1st term has power as -1/3. u hv written 1/3.  put f'(x) u got equal to 0.

[Dreyfus]

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[SoniaKapoor]

i maen the critical values is x=3 and f '(x) is not defined at x=0...so we wont check at x=0???

[Dreyfus]

Oh I missed that ......f(0) is 0.....so it can't be maximum value......

[Homer Simpson]

f(5) is mentioned as 4.something - so f(0) = 0 wont work anyway as maximum. f(3), f(-1), f(5) are the other possibilities.

[SoniaKapoor]

i knw it wont work ..my quest is we have to check where f(x) is not defined also?at 0 in this case

[Dreyfus]

Yes func can attain extrema at kink also.......

[SoniaKapoor]

ok then -1,0,3 and 5 are the pts at which we check
but fn does not change sign at 5...so ruled out
am i right vaibhav?

[Dreyfus]

Actually this function is non negative throughout its domain.....so u hv to evaluate values at each of these points,...that's  y in the ques f(5) is given as it is too tedious to find f(5) in the exam.....

[SoniaKapoor]

thank you...