For every value of x,the function f(x)=frac{1 | vedclass

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(A) Given the function $f(x) = \frac{1}{a^x} = a^{-x}$,where $a > 0$.To determine the nature of the function,we find its derivative with respect to $x

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decreasing

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(A) Given the function $f(x) = \frac{1}{a^x} = a^{-x}$,where $a > 0$.To determine the nature of the function,we find its derivative with respect to $x$:$f'(x) = \frac{d}{dx}(a^{-x}) = -a^{-x} \cdot \ln(a)$.Since $a^x > 0$ for all $x$ and $a > 0$,the sign of $f'(x)$ depends on the value of $a$:$1$. If $a > 1$,then $\ln(a) > 0$,which implies $f'(x) $2$. If $0  0$,so the function is strictly increasing.$3$. If $a = 1$,then $f(x) = 1$,which is a constant function.Since the question asks for the nature of the function for every value of $x$ and $a > 0$ (without specifying $a > 1$),the function is not strictly increasing or decreasing for all $a$. However,in standard textbook contexts where $a$ is assumed to be a base such that $a > 1$,it is decreasing. Given the options,if $a > 1$,the function is decreasing.

For every value of $x$,the function $f(x)=\frac{1}{a^{x}}, a>0$ is

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