# Homework Help: Exponential Growth Functions

1. Jan 5, 2012

### darshanpatel

1. The problem statement, all variables and given/known data

Consider the exponential function f(x)= ab^(x)

a) Show that (f(x+1))/f(x) = b

b)Use the results from part (a) to explain why there is no exponential function of the form f(x)= ab^(x) whose graph passes through the points (0,4) (1,4) (2,8) (3,24) and (4,72)

2. Relevant equations

-None-

3. The attempt at a solution

There is no attempt at part (a) becuase I couldn't find a way to start it, but for part (b) I am thinking it can't be a exponential function becuase the y-values for two of the x-values are the same. That's what I think, except I can't put it togethor the way part (b) wants me to.

2. Jan 5, 2012

### eumyang

You are given
$f(x) = ab^x$
Are you familiar with function notation? What goes inside the parentheses next to f can indicate what you're plugging in. For example,
$f(5) = ab^5$
$f(\text{Mickey Mouse}) = ab^\text{Mickey Mouse}$
Given that, what is
$f(x+1) ?$
Once you find that, then substitute into
$$\frac{f(x+1)}{f(x)}$$
Suppose (0, 4) is on the graph of f(x). This is saying that
$f(0) = 4$
Suppose the other points are on the graph of f(x) as well. What is
$f(1) ?$
$f(2) ?$
...
Can you see how this relates to the expression
$$\frac{f(x+1)}{f(x)} ?$$

3. Jan 5, 2012

### darshanpatel

Thanks, I know get part (a), I was misunderstanding the problem before. It has to be like (ab^(x+1))/(ab^(x)) and then you show simplification. :) :)

Sorry about that whole thing, I understand it now, but how do I take this and put it with part (b)?

Last edited: Jan 5, 2012
4. Jan 5, 2012

### eumyang

If you mean
$$\frac{ab^x + 1}{ab^x}$$
... then sorry, that's not right. Is the "+1" really supposed to be separate from abx?

5. Jan 5, 2012

### eumyang

In (b), in order for the points to lie on the graph of f(x) = abx, then for any x,
$\frac{f(x+1)}{f(x)}$
has to equal the same base b. So what is
$\frac{f(1)}{f(0)} ?$
$\frac{f(2)}{f(1)} ?$
And so on.

6. Jan 5, 2012

### darshanpatel

I don't understand what you are saying...

The question:

b)Use the results from part (a) to explain why there is no exponential function of the form f(x)= ab^(x) whose graph passes through the points (0,4) (1,4) (2,8) (3,24) and (4,72)

From the points, which is the base? I understand you take the points and plug it into the equation, but I don't get anything further then that. For example, for the point (0,4). I would do this:

y=b^x
4=b^0
4≠1

Is that right to show that it is not a function becuase the values are unequal.

^^Thats probably wrong but what do I plug into the (f(x+1))/(f(x)) =b ????? <--- Thats where I am getting confused

Last edited: Jan 5, 2012
7. Jan 5, 2012

### darshanpatel

bumping this becuase its the last problem and I am getting irritated :(

8. Jan 6, 2012

### eumyang

Why did you plug into y = bx? The function is in the form of y = abx. Anyway, that's not what I would do.

Let's look at another exponential function: y = 5*2x. Some of the points are these:
(0, 5), (1, 10), (2, 20), (3, 40), (4, 80)...
For an exponential function f, and for any x,
$\frac{f(x+1)}{f(x)} = b$.
Take the ratio of successive y-values, like thus:
$\frac{f(1)}{f(0)} = \frac{10}{5} = 2$
$\frac{f(2)}{f(1)} = \frac{20}{10} = 2$
$\frac{f(3)}{f(2)} = \frac{40}{20} = 2$
$\frac{f(4)}{f(3)} = \frac{80}{40} = 2$
...
Note that they all equal 2, and that is the base of the exponential function y = 5*2x.

Now look at the points you were given:
(0,4) (1,4) (2,8) (3,24) and (4,72)
and find
$\frac{f(1)}{f(0)} = ?$
$\frac{f(2)}{f(1)} = ?$
$\frac{f(3)}{f(2)} = ?$
$\frac{f(4)}{f(3)} = ?$
Do these simplify to the same number?