which function has the same range as $f(x)=-\frac{5}{7}(\frac{3}{5})^x$?\n$g(x)=\frac{5}{7}(\frac{3}{5})^{-x}…

Answer:

B. $g(x)=-\frac{5}{7}(\frac{3}{5})^{-x}$

Explanation:

Step1: Analyze the range of $f(x)$

For the exponential - function $y = a\cdot b^{x}$, when $b\in(0,1)$ and $a\lt0$, as $x\rightarrow+\infty$, $b^{x}\rightarrow0$ and $y\rightarrow0$ (from the negative side), and as $x\rightarrow-\infty$, $b^{x}\rightarrow+\infty$ and $y\rightarrow-\infty$. For $f(x)=-\frac{5}{7}(\frac{3}{5})^{x}$, since $a =-\frac{5}{7}\lt0$ and $b=\frac{3}{5}\in(0,1)$, the range of $f(x)$ is $(-\infty,0)$.

Step2: Analyze option A

For $g(x)=\frac{5}{7}(\frac{3}{5})^{-x}=\frac{5}{7}(\frac{5}{3})^{x}$, since $a=\frac{5}{7}\gt0$ and $b = \frac{5}{3}\gt1$, the range is $(0,+\infty)$.

Step3: Analyze option B

For $g(x)=-\frac{5}{7}(\frac{3}{5})^{-x}=-\frac{5}{7}(\frac{5}{3})^{x}$, since $a =-\frac{5}{7}\lt0$ and $b=\frac{5}{3}\gt1$, as $x\rightarrow+\infty$, $(\frac{5}{3})^{x}\rightarrow+\infty$ and $g(x)\rightarrow-\infty$, as $x\rightarrow-\infty$, $(\frac{5}{3})^{x}\rightarrow0$ and $g(x)\rightarrow0$ (from the negative side). The range of $g(x)$ is $(-\infty,0)$.

Step4: Analyze option C

For $g(x)=\frac{5}{7}(\frac{3}{5})^{x}$, since $a=\frac{5}{7}\gt0$ and $b=\frac{3}{5}\in(0,1)$, the range is $(0,+\infty)$.

Step5: Analyze option D

For $g(x)=-(-\frac{5}{7})(\frac{5}{3})^{x}=\frac{5}{7}(\frac{5}{3})^{x}$, since $a=\frac{5}{7}\gt0$ and $b=\frac{5}{3}\gt1$, the range is $(0,+\infty)$.