The function
is decaying.
Explanation:
Given:
![f(x)=\left((1)/(5)\right)^(x)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/cx1v74p6qr0idyvqpwlus0bzi9613j0kh5.png)
The above equation can also write as below,
![f(x)=(1)/(5^(x))](https://img.qammunity.org/2020/formulas/mathematics/middle-school/cntcr4r5f082v4mx712ldz7sdjror3rmek.png)
Now from the function we can see that as we are increasing the value of x the value of function is decreasing. Let take some values of x and then find out the value of function at those value of x
When x = 1 ,
![f(1)=(1)/(5)=0.2](https://img.qammunity.org/2020/formulas/mathematics/middle-school/adbn6mgegpgfu6bfunqrwe9vtlt4sdcc2n.png)
When x = 2,
![f(2)=(1)/(5^(2))=(1)/(25)=0.04](https://img.qammunity.org/2020/formulas/mathematics/middle-school/2fxo3m686u6m9jwaa6s65receew49swm1r.png)
When x = 3,
![f(3)=(1)/(5^(3))=0.008](https://img.qammunity.org/2020/formulas/mathematics/middle-school/s7rwq6593e99n0ogm7jzclznfv3fiuul31.png)
So, from the above examples, we can see that the value of function is decreasing on increasing the value of x. Hence the function is decay