If x = –0.5, then which of the following has the smallest value?

$x=\frac{-1}{2}$

$Option(a):2^{\frac{1}{x}}=2^{\frac{1}{\frac{-1}{2}}}=2^{-2}=\frac{1}{4}$

$Option(b):\frac{1}{x}=\frac{1}{\frac{-1}{2}}=-2$

$Option(c):\frac{1}{x^2}=\frac{1}{{\left(\frac{-1}{2}\right)}^2}=4$

$Option(d):2^x=2^{\frac{-1}{2}}=\frac{1}{2^{\frac{1}{2}}}=\frac{1}{\sqrt{2}}$

$\text{So, smallest is }\frac{1}{x}.$

By looking at the options, we can find that the value given in option (b) will be negative and remaining options will give us a positive value.

So, the smallest value will be $\frac{1}{x}$.