## anonymous one year ago Task 1 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.

1. anonymous

@peachpi I need you when you get a chance

2. anonymous

|dw:1437340736543:dw|

3. anonymous

can I insert any values

4. anonymous

yes you can use any value. I was asking which one of those is the right format for f(x)

5. anonymous

f(x)=x+a/b

6. anonymous

I'm guessing you mean the top one? If that's the case, b = c and a = d.

7. anonymous

so you only need to pick 2 numbers

8. anonymous

ok what about 7 and 5

9. anonymous

Actually that doesn't make sense because then you could just subtract a/b. Is just a in the numerator of the fraction?

10. anonymous

yes so are there any specific numbers I have to put In because remember f(x) ad g(f) have to be inverses

11. anonymous

tbh I'm kind of confused. I'm still stuck on what f(x) is supposed to be. The format makes a difference because of order of operations. Is it the top one I have drawn above, or the bottom?

12. anonymous

its the bottom one

13. anonymous

ok. so then yes you can use 7 and 5.

14. anonymous

which one is a and which is b?

15. anonymous

so I fill in 7 and 5 for x or a b

16. anonymous

$f(x)=\frac{ x+7 }{ 5 }$

17. anonymous

So then the inverse would be g(x)=5x - 7

18. anonymous

ok now how would I show work to prove that the inverse of f(x) is g(x)

19. anonymous

are there specific steps that I have to go through

20. anonymous

Find f(g(x)) or g(f(x)). If you get x, they're inverses

21. anonymous

ok so when I graph them how would I do that

22. anonymous

https://www.desmos.com/calculator you can use that to graph. just use y= in front of the equation

23. anonymous

ok thanks for all your help today sorry to ask so many questions

24. anonymous

no problem. good luck with your class :)