Question

How do you tell whether `f(x)=(6/5)^-x` is an exponential growth or decay?

Answer 1

Given the form `f(x)=A^-x`

If `A>1` `->` exponential decay
If `A<1` `->` exponential growth

Explanation:

To understand why this makes sense, let's expand the function that is given. To do this, we need to keep in mind two properties of exponents:

First, `(a/b)^x=a^x/b^x`

Second, `a^-x=1/a^x`

Using the first property, expand the equation given in the problem:

`f(x)=(6/5)^-x=6^-x/5^-x`

Now, use the second property to flip the fraction and drop the negative sign from the exponent:

`f(x)=5^x/6^x`

Now that we've gotten this far, let's consider what will happen to the value of `f(x)` as `x->oo` by looking at the graphs of `5^x` and `6^x`.

Graph of `5^x`
graph{5^x [-5, 5, -1.47, 5]}

Graph of `6^x`
graph{6^x[-5, 5, -1.47, 5]}

Looking at the difference between these two graphs, `6^x` is steeper than `5^x`. This implies that the denominator of `f(x)=5^x/6^x` will grow faster, making `f(x)` smaller as `x` increases (exponential decay).

If `6^x` were in the numerator, the opposite would be true, and we would see exponential growth.

Answer 2

Decay

Explanation:

Given:`" "f(x)=(6/5)^(-x)`

Set as:`y=(6/5)^(-x)`

The index (power) being negative is another way of writing:

`y=(color(white)(.)1color(white)(.))/(6/5)^x`

This is the same as:

`y=(5/6)^x" "=5^x/6^x`

As `x` increases then `6^x` becomes increasingly greater that `5^x`

Thus the whole will become less and less as `x` becomes greater and greater eventually approaching a particular value.

Thus we have `y=5^x/6^x->k` as `x` increases

And `lim_(x->oo)y=lim_(x->oo)k=0`

Thus it is decay