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.
plz help
@zepdrix
@undeadknight26
plz help
@Here_to_Help15
will fan you if i know how to
will fan you
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help
Ok
thx
@Here_to_Help15 are you there
We want f(x) and g(x) to be inverse functions of each other. In general, if you want to find the inverse function, rearrange the equation so that you solve for x, instead of f(x), then make every x an f(x) and make every f(x) an x (i.e. switch the variables). What you're left with is the inverse of what you started with. So...choose any two numbers you want for a and b, then rearrange the equation as I described above. Look at your new equation, and if you put it in the same form as g(x) above, you'll see what c and d are right away!
ok
you mean like x=f(x)+ab
\[f(x)=x+a/b\]
hello
I mean solve for the other variable, THEN switch. So if you're starting with: \[f(x)=x+{a \over b}\] Solving for x instead of f(x) gives you: \[x=f(x)-{a \over b}\] But this is still the same FUNCTION as you started with, because all you did was rearrange the terms. Your INVERSE function will be in the same form, but with x and f(x) switched - this creates a new function because you have changed the relationship between x and f(x): \[f^{-1}(x)=x-{a \over b}\]
ok thx
can you help me with another parts too
\[f(x)=(x+a)/b\]
the equation wasn't f(x)=x+a/b
it was the whole x+a /b
Ah gotcha Well same process for that function. What is the inverse of that?
x=(f(x)+a)/b?
and sorry about this mess i am new to this program
That's ok! So yes, that would be the inverse function, but now isolate f(x) so that it's alone on the left side of the equation. What do you get?
is it f(x) = x/b +a/b
hold on
\[(x+a)/b=f(x)?\]
Not quite. Look: \[x= {f(x)+a \over b}\]\[x \times b = {f(x)+a \over b} \times b\]\[bx= f(x)+a \]\[bx-a= f(x)+a -a\]\[f(x)=bx-a \] So this is our inverse function. The question calls this g(x), with variables c and d. Looking at this new equation, what does c equal and what does d equal?
yes
i think so
matt101 you there?
Yes I am - just waiting for you to answer my question!
i said yes i think so
yes it does equal c and d
matt101 you there
matt 101 you there?
@matt101
hello anyone?
matt 101 are you there
Yup sorry just stepped away for a second. So what exactly does c equal? And what exactly does d equal?
sorry i was off too
c is the y-intercept and d is slope?
you there matt101
Well it's the other way around actually... But I was referring to the actual values of c and d. Have a look at the inverse function I got above, and what they give as g(x) in the question: f(x)=bx-a g(x)=cx-d Soooo what does c equal and what does d equal?
c is y-intercept and d is slope?
What I'm trying to get you to say is that c=b and d=a. Look at the two equations I wrote above. And by the way, when you have an equation in y=mx+b form, the m is the slope (in this case c) and the b is the y-intercept (in this case d).
so sorry i am confused
ok
matt101 you there
Yes. What don't you understand?
oh never mind now i get it
that is it for part A b,c,d
Well no, for part A you need numbers for a, b, c, and d. Choose any two numbers you want. You only need two because a=d and b=c. For part B, write in these numbers instead of a, b, c, and d, and find the inverse of f(x) to show that it's equal to g(x)
how about 5 and 2
Sure
which means we have to insert it for c and d
Right
which means that is it for part 1 and part 2
Yes. Now for part C - you need to find g(f(x)). That means in g(x), you need to sub in f(x) for every x.
what did mean by you need to sub in f(x) for every x i didn't get that
do you mean g(f(x))
Yes. g(x)=cx-d Or in your case, g(x) = 5x-2 Now, g(f(x))=5f(x)-2 We know f(x)=(x+2)/5, so our equation then becomes: g(f(x))=5((x+2)/5)-2 So I've pretty much given you the answer for part C now lol
But you can simplify that last equation just a bit further...and you might be surprised at what you get!
ok
is it g(f(x))=x
Absolutely right!
That's it for part C. For the last part you just need to graph f(x) and g(x) and the line y=x. Come up with a table of values for each to make them easy to plot!
hold on
so basically we are just making a table out of random choice of ordered pairs in x and y
Yup
domain means x right?
domain: 5,6,9,4,8 range: 1,4,3,2,7
is it like that
For which function?
g(x)?
If you plug in x=5 to g(x) what do you get?
g(5)?
Yes what is g(5)?
i am confused at part d
You're just graphing the functions the way you would for any other. If you sub 5 into g(x)=5x-2 what do you get?
i am weak at algebra
g(5)=5(5)-2 g(5)=?
g(5)=25-2 g(5)=23
Right! That's one point on your graph for g(x) - (5,23). Now repeat the process for 4 other values of x. Use the 5 points to draw the line. Then do the same for f(x), and once more for y=x.
oh now i get it
Great! Glad I could help :)
g(6)=28
g(9)=43
g(4)=18
g(8)=38
is it correct?
Yup
now the ordered pair for g(6) would look like this (6,28)
@matt101
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