9. f(x) = (1+x)^0.3/1+x^0.3 in the interval [0,1] is attained at x equal to:
A. 1
B. 2^0.7
C. 2^-0.7
D. 0
Ans. A
Please give explanation???
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The minimum of this function on #[0,1]# is #f(1)=\frac{2^0.3}{2}# , using the Closed Interval Method.
The question doesn't seem complete, but I imagine you are looking for extrema of the function #f(x) = (1+x)^0.3/(1+x^0.3)# on #[0,1]#, specifically, where the absolute minimum is reached. We know (Closed Interval Method ) that these may occur at the end points, or at critical values inside the interval #(0,1)#.
Since #f'(x)=\frac{0.3(1+x)^{-0.7}(1+x^0.3)-0.3x^(-0.7)(1+x)^0.3}{(1+x^0.3)^2}#
using the Quotient Rule , we see that #f'(x)=0# if #(1+x)^{-0.7}(1+x^0.3)-x^(-0.7)(1+x)^0.3=\frac{1+x^0.3}{(1+x)^0.7}-\frac{(1+x)^0.3}{x^0.7}=0#, that is, if
#x^0.7(1+x^0.3)-(1+x)^0.7(1+x)^0.3=0#, equivalently, if
#x^0.7+x-(1+x)=0#, that is, if
#x^0.7=1#, equivalently, #x=1#.
In other words, there is no critical point in #(0,1)# and extrema can only occur at 0 and 1. Moreover, #f(0)=1# and #f(1)=\frac{2^0.3}{2}\approx 0.6#.
