Question

Let $h\left(x\right)=x^{\frac{m}{n}}$ for $x\in R$, where $m$ and $n$ are odd numbers and $0<m<n$, then $y=h\left(x\right)$ has

• A. No local extremum
• B. One local maximum
• C. One local minimum
• D. None of the above

Answer 1

The correct option is A No local extremum

The given equation is:

$y=x^{\frac{m}{n}}$

To find the extremum points we differentiate and equate it to zero

$\Rightarrow y^{\prime }=\frac{m}{n}\times x^{\frac{m}{n}-1}$

$y^{\prime }=0$

$\Rightarrow \frac{m}{n}\times x^{\frac{m}{n}-1}=0$

$\Rightarrow x^{\frac{m}{n}-1}=0$

$0<m<n$

$\Rightarrow \frac{m}{n}<1$

$\Rightarrow \frac{1}{x^{1-\frac{m}{n}}}=0$.

For this to be true we have to make $x\to \infty$ which suggest of no extremum values of the function y. Hence option A is correct.