Graphical Analysis of Functions: Finding g(x+3) at x=1

[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?

------------------

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)?

 

[coburg]

listen up sportsguy...

ur first working is correct...

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

 

[coburg]

listen up sportsguy...

ur first working is correct...

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

 

[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)...

 

[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).

 

[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).

 

[sportsguy3675]

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