Operations of Functions

  • #1
Little help please and some work checking.

Given [tex]f(x)=3x+2[/tex] and [tex]g(x)=\frac{x-4}{2x}[/tex]

It asks for [tex]g(\frac{1}{x})[/tex]

So substitute in: [tex]\frac{\frac{1}{x} - 4}{2(\frac{1}{x})}[/tex]

Simplify: [tex]\frac{\frac{1}{x} - 4}{\frac{2}{x}}[/tex]

I multiply top and bottom by x right?

That would give: [tex]\frac{1 - 4x}{2}[/tex] and [tex]D_{f} = \Re[/tex] 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)?
 

Answers and Replies

  • #2
7
0
listen up sportsguy....

ur first working is correct....

for the second one...do u hav only the graph or is the equatios provided?
 
  • #3
7
0
listen up sportsguy....

ur first working is correct....

for the second one...do u hav only the graph or is the equatios provided?
 
  • #4
7
0
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
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
HallsofIvy
Science Advisor
Homework Helper
41,833
956
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
Yeah, but see it didn't say If, it said at. But I did look on the graph for f(4). :)
 

Related Threads on Operations of Functions

Replies
1
Views
3K
Replies
1
Views
936
Top