# Symmetricity of exponential graphs

1. Nov 13, 2008

### l46kok

1. The problem statement, all variables and given/known data
Suppose two exponential functions, f(x) and g(x) are symmetric with respect to x = 2.

$$f(x)=a^{bx-1}$$
$$g(x)=a^{1-bx}$$

Prove f(2) = g(2)

2. Relevant equations

3. The attempt at a solution
This isn't actually a problem. This is a property used to solve a different problem which I really had a hard time understanding. Why is this true?

Last edited: Nov 13, 2008
2. Nov 13, 2008

### Staff: Mentor

I might be completely misunderstanding what you're trying to say, but the functions you give are not symmetric with respect to the line x = 2, and furthermore, f(2) $$\neq$$ g(2). The two functions are reciprocals of one another, though.

$$f(2) = a^{2b - 1}$$
$$g(2) = a^{1 - 2b} = a^{-(2b -1)} = \frac{1}{a^{2b - 1}} = \frac{1}{f(2)}$$

If the two functions were the mirror images of each other across the line x = 2, it would have to be the case that f(2 + x) = g(2 - x). For example, f(3) would have to be equal to g(1), which is not the case.

Can you clarify what you're trying to do?

3. Nov 13, 2008

### l46kok

Well, here's the whole problem:

The two exponential functions

$$f(x)=a^{bx-1}$$
$$g(x)=a^{1-bx}$$

Satisfy the following conditions

a)Functions y = f(x) and y=g(x) are symmetric to the line x = 2 when graphed
b)$$f(4) + g(4) = \frac {5}{2}$$

Find a and b.

4. Nov 13, 2008

### Dick

a^(bx-1) is not symmetric with respect to any x value unless either b=0 or a=1. a=1 doesn't work if f(4)+g(4)=5/2. Hence b=0. So a+a^(-1)=5/2. That's easy to solve. Is this problem as silly as that?

5. Nov 14, 2008

### borgwal

Perhaps the question is intended to mean

f(2-x)=g(x-2) for all x.

6. Nov 14, 2008

### Dick

So b(2-x)-1=1-b(x-2) for all x? About all I get out of that is 1=(-1).

7. Nov 14, 2008

### borgwal

Sorry, I meant of course that the condition is

f(2+x)=g(2-x) for all x.

8. Nov 14, 2008

### Dick

No, I'm sorry. I knew what you meant to say, but I didn't check that f(2-x)=g(x-2) was not the symmetry we were both thinking of. Yeah, that seems to give a reasonable answer for a and b. l46kok, look at how I tried to derive the value of b incorrectly and use borgwal's corrected condition on f and g to get b.

9. Nov 25, 2008

### l46kok

Something is wrong, the answer is supposed to be

a+b = 1 though

10. Nov 25, 2008

### l46kok

I asked my professor and he told me that

g(x) = f(4 - x), f(x) = g(4 - x), since x is symmetric to 2

Now I'm even more confused.

11. Nov 25, 2008

### Dick

Why don't you try substituting that condition into the definitions of f(x) and g(x)?

12. Nov 25, 2008

### Chaos2009

g(x) = f(4 - x), f(x) = g(4 - x), and f(x - 2) = g(x + 2) actually all say the same thing in a slightly different way.

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