For every value of $x$, the function $f(x)=\frac{1}{a^{x}}, a0$ is

Question

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

Aicharya · Accepted Answer

$\begin{aligned} \mathrm{f}(\mathrm{x})=& \frac{1}{\mathrm{a}^{\mathrm{x}}}=\mathrm{a}^{-\mathrm{x}} \ 	herefore \quad \mathrm{f}^{\prime}(\mathrm{x}) &=-\mathrm{a}^{-\mathrm{x}} \cdot \log _{e} \mathrm{a}=-\frac{\log _{\mathrm{e}} \mathrm{a}}{\mathrm{a}^{\mathrm{x}}}= < 0 \end{aligned}$ So $\mathrm{f}(\mathrm{x})$ is decreasing for all $\mathrm{x}$.

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

Solution