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The maximum value of[x(x-1)+1]^(1/3), 0 le x le1 is: |
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Answer» Solution :Let ` f(x) = [x(x-1)+1]^(1//3)=(x^(2)-x+1)^(1//3),0 LEX le1` Differentiate w.r.t.X, `f'(x) = 1/3 (x^(2)-x+1)^(1/3-1)(2x-1)=((2x-1))/(3(x^(2)-x+1)^(2/3))` f'(x) = 0, ` rArr2x - 1 = 0rArrx=1/2 in [0, 1]` Now, we find thevalues of f at `x= 1/2` and at the end points of the interval [0, 1] at ` x= 0,f(0) = (0-0+1)^(1//3) = 1` at ` x = 1, f(1) = (1-1+1)^(1//3) = 1` at ` x = 1/2,f(1/2)=(1/4-1/2+1)^(1/3) = (3/4)^(1/3)` `:. ` The maximum VALUE off(x) is, 1, at x = 0, 1. |
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