# Homework Help: Operations of Functions

1. Aug 30, 2005

### sportsguy3675

Little help please and some work checking.

Given $$f(x)=3x+2$$ and $$g(x)=\frac{x-4}{2x}$$

It asks for $$g(\frac{1}{x})$$

So substitute in: $$\frac{\frac{1}{x} - 4}{2(\frac{1}{x})}$$

Simplify: $$\frac{\frac{1}{x} - 4}{\frac{2}{x}}$$

I multiply top and bottom by x right?

That would give: $$\frac{1 - 4x}{2}$$ and $$D_{f} = \Re$$ Correct?

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Then on this other problem I have a graph of different functions f and g. The problem says g(x+3) at x=1. Does that mean I find the y value on g where x = 1 (4) and then add 3 and then find g(7)? Or do I add 1 to 3 and take g(4)?

2. Aug 30, 2005

### coburg

listen up sportsguy....

ur first working is correct....

for the second one...do u hav only the graph or is the equatios provided?

3. Aug 30, 2005

### coburg

listen up sportsguy....

ur first working is correct....

for the second one...do u hav only the graph or is the equatios provided?

4. Aug 30, 2005

### coburg

its pretty simple for the second one too where in case u were jus given the graphs.....
g(x+3) means notin but shifting the graph by 3 units to the left.....if u could redraw the graph then....just draw it and look for the value at one....or the more simpler approach would be to jus look for the value at g(4)......

5. Aug 30, 2005

### sportsguy3675

OK, thanks.

On the 2nd part, it has nothing to do with graphing. There is just a graph with f and g drawn on it and you just have to find the value based on that graph. All the other questions were normal but I don't understand the whole at x = whatever business. That why I asked if it was asking for g(4) or g(7).

6. Aug 31, 2005

### HallsofIvy

The second problem says f(x+3)- that is x+ 3, not y+ 3! Also, that "x+3" is inside the parentheses for the function.

If x= 1 then you add x not y: x+ 3= 1+ 3= 4. Then apply the function.
If x= 1, f(x+3)= f(1+ 3)= f(4).

7. Aug 31, 2005

### sportsguy3675

Yeah, but see it didn't say If, it said at. But I did look on the graph for f(4). :)