# 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???

Jun 11, 2018

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

#### Explanation:

The question doesn't seem complete, but I imagine you are looking for extrema of the function $f \left(x\right) = {\left(1 + x\right)}^{0.3} / \left(1 + {x}^{0.3}\right)$ on $\left[0 , 1\right]$, 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 $\left(0 , 1\right)$.

Since $f ' \left(x\right) = \setminus \frac{0.3 {\left(1 + x\right)}^{- 0.7} \left(1 + {x}^{0.3}\right) - 0.3 {x}^{- 0.7} {\left(1 + x\right)}^{0.3}}{{\left(1 + {x}^{0.3}\right)}^{2}}$
using the Quotient Rule , we see that $f ' \left(x\right) = 0$ if ${\left(1 + x\right)}^{- 0.7} \left(1 + {x}^{0.3}\right) - {x}^{- 0.7} {\left(1 + x\right)}^{0.3} = \setminus \frac{1 + {x}^{0.3}}{{\left(1 + x\right)}^{0.7}} - \setminus \frac{{\left(1 + x\right)}^{0.3}}{{x}^{0.7}} = 0$, that is, if
${x}^{0.7} \left(1 + {x}^{0.3}\right) - {\left(1 + x\right)}^{0.7} {\left(1 + x\right)}^{0.3} = 0$, equivalently, if
${x}^{0.7} + x - \left(1 + x\right) = 0$, that is, if
${x}^{0.7} = 1$, equivalently, $x = 1$.

In other words, there is no critical point in $\left(0 , 1\right)$ and extrema can only occur at 0 and 1. Moreover, $f \left(0\right) = 1$ and $f \left(1\right) = \setminus \frac{{2}^{0.3}}{2} \setminus \approx 0.6$.