Let m and n be odd integers such that 0

Question

(D) Given $f(x) = x^{\frac{m}{n}}$,where $m$ and $n$ are odd integers and $0 < m < n$.First,consider the derivative: $f'(x) = \frac{m}{n} x^{\frac{m}{

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$f$ increases on $\mathbb{R}$.

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(D) Given $f(x) = x^{\frac{m}{n}}$,where $m$ and $n$ are odd integers and $0 First,consider the derivative: $f'(x) = \frac{m}{n} x^{\frac{m}{n} - 1} = \frac{m}{n} x^{\frac{m-n}{n}}$.Since $m As $x 	o 0$,$f'(x) 	o \infty$,so $f'(0)$ does not exist. Thus,$f(x)$ is not differentiable at $x = 0$.Since $k$ is even,$x^{\frac{k}{n}} > 0$ for all $x 
eq 0$. Therefore,$f'(x) = \frac{m}{n \cdot x^{\frac{k}{n}}} > 0$ for all $x 
eq 0$.Since $f'(x) > 0$ for all $x \in \mathbb{R} \setminus \{0\}$ and $f(x)$ is continuous at $x = 0$,the function $f(x)$ is strictly increasing on the entire domain $\mathbb{R}$.Thus,the correct option is $(D)$.

Let $m$ and $n$ be odd integers such that $0 < m < n$. If $f(x) = x^{\frac{m}{n}}$ for $x \in \mathbb{R}$,then:

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