horsegirl27
  • horsegirl27
Please help me out!! I have work for the first part of this question, but then I'm confused. When reading it, it seems like parts 2 and 3 are the same, so I would like some clarification of what to do. For part 4, I'm not quite sure either.... Question posted below.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
horsegirl27
  • horsegirl27
Part 1. Using the two functions listed below, insert numbers in place of the letters a, b, c, and d so that f(x) and g(x) are inverses. f(x)= x+a b g(x)=cx−d Part 2. Show your work to prove that the inverse of f(x) is g(x). Part 3. Show your work to evaluate g(f(x)). Part 4. Graph your two functions on a coordinate plane. Include a table of values for each function. Include 5 values for each function. Graph the line y = x on the same graph.
horsegirl27
  • horsegirl27
I'm sorry, those functions are messed up, let me fix that.
horsegirl27
  • horsegirl27
f(x)=x+a/b g(x)=cx-d

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

horsegirl27
  • horsegirl27
So, for part 1, I have f(x)=x+3/5 and g(x)=5x-3. I'm not sure of what I'm doing in parts 2 and 3.
phi
  • phi
is f(x) \[ f(x) = \frac{x+a}{b} \]?
horsegirl27
  • horsegirl27
yes.
phi
  • phi
so we could write \[ f(x) = \frac{1}{b} x + \frac{a}{b} \]
horsegirl27
  • horsegirl27
For which part?
phi
  • phi
just thinking out loud. For part 2, they want you to show the functions are inverses. start with your: f(x)=(x+3)/5 g(x)= 5x-3 if they are inverses then f( g(x) ) = x
phi
  • phi
f( g(x) ) is short-hand for : everywhere you see "x" in the expression for f(x), erase the x and write g(x) using f(x)=(x+3)/5 g(x)= 5x-3 f( g(x) ) = ( g(x) +3 ) /5
horsegirl27
  • horsegirl27
ok
phi
  • phi
now we use g(x) = 5x-3 in f( g(x) ) = ( g(x) +3 ) /5 in other words, on the right hand side, replace g(x) with its "definition" 5x-3 can you do that ?
phi
  • phi
you should get \[ f( g(x) ) = \frac{( 5x-3 +3)}{5} \]
phi
  • phi
now simplify that. For part 3, do almost the same thing show g( f(x) ) = x
horsegirl27
  • horsegirl27
ok
horsegirl27
  • horsegirl27
And I am so sorry, I was gone for a moment.
horsegirl27
  • horsegirl27
Ok, so that's what I got for that part.
horsegirl27
  • horsegirl27
So for part 3, to I basically just solve it to show g( f(x) ) = x?
horsegirl27
  • horsegirl27
Because I worked it out and got it. Is that what I want to turn in for part 3, my work?
phi
  • phi
I assume for part 2, you got f( g(x))= x after simplifying?
phi
  • phi
\[ f( g(x) ) = \frac{( 5x-3 +3)}{5} = \frac{5x+0}{5} = \frac{5x}{5} = x \]
horsegirl27
  • horsegirl27
yes. That's my work exactly.
phi
  • phi
for part 3 g( f(x) ) start with f(x)=(x+3)/5 g(x)= 5x-3 g(f(x)) means replace x in 5x-3 with f(x) what do you get ?
horsegirl27
  • horsegirl27
g(f(x))=5(x+3/5)-3 Then the 3's will cancel each other out, and so will the 5's, to get g(f(x))=x.
phi
  • phi
you have to remember order of operations. \[ g(f(x)) = 5 \cdot \frac{(x+3)}{5} -3 \] multiply before add/subtract
horsegirl27
  • horsegirl27
oh, sorry
horsegirl27
  • horsegirl27
So then it will x+3/1 or 0? And then after that, the 3's will still cancel out, right?
phi
  • phi
the 5/5 becomes 1 and you get \[ g(f(x)) = 5 \cdot \frac{(x+3)}{5} -3 \\ = 1\cdot (x+3) -3 \\=\ (x+3)-3 \\ =x+3-3 \\=x\]
phi
  • phi
just like \[ \frac{15}{5} = \frac{5 \cdot 3 }{5} = \frac{5}{5} \cdot 3 = 1 \cdot 3 = 3\]
horsegirl27
  • horsegirl27
ok, that makes sense.
horsegirl27
  • horsegirl27
So for part 2, I'm showing for for f(x), and for part 3, showing work for g(x)? Then graphing? And I'm confused how I should graph this
phi
  • phi
Include a table of values for each function. Include 5 values for each function. that means they want a table of x y values where y is f(x) or g(x) depending on which function you are doing.
horsegirl27
  • horsegirl27
ok, so for x I could use something like 0, 1, 2, 3, -1 and for y -3, 2, 7, 12, -8? (i might not use those numbers, I just chose them randomly)
phi
  • phi
Start with the first function f(x) f(x)=(x+3)/5 I would pick small negative and positive numbers for x to get integer answers for f(x) (i.e. y) for example if you start at x= -8, and make x go up by 5 , you will get nice numbers for f(x)
horsegirl27
  • horsegirl27
ok, thanks.
horsegirl27
  • horsegirl27
By the way, any suggestions for an online graph?
phi
  • phi
I downloaded geogebra.
horsegirl27
  • horsegirl27
Okay, I'll download it too.
horsegirl27
  • horsegirl27
So, graph 5 points for each?
phi
  • phi
First write down the table x | f(x) -------- -8 | -1 -3 | 0 etc and for g(x) you can pick x= -2 to +2 by 1 x | g(x) -------- -2 | -13 -1 | -8 etc
phi
  • phi
if you use geogebra you type in f(x)= (x+3)/5 and it will plot the line
horsegirl27
  • horsegirl27
ok
horsegirl27
  • horsegirl27
Got it, thanks!
horsegirl27
  • horsegirl27
I have one more question, want me to make a new thread?
phi
  • phi
Here is the plot
1 Attachment
horsegirl27
  • horsegirl27
thank you :D

Looking for something else?

Not the answer you are looking for? Search for more explanations.