Question

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???

Answer 1

The minimum of this function on `[0,1]` is `f(1)=\frac{2^0.3}{2}` , using the Closed Interval Method.

Explanation:

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`.