## anonymous one year ago Will METAL!! Confirm that f and g are inverses by showing that f(g(x)) = x and g(f(x)) = x. f(x) = the quantity x minus seven divided by the quantity x plus three. and g(x) = quantity negative three x minus seven divided by quantity x minus one.

1. phi

ooh. Time to use the equation editor.

2. anonymous

|dw:1443790834332:dw|

3. anonymous

@phi finished

4. phi

the first part f( g(x) ) means everywhere you see x in f(x), replace it with g(x)

5. anonymous

how exactly do I replace it.

6. imqwerty

Metal (:

7. phi

the first step is $f( g(x) ) = \frac{g(x) -7}{g(x)+3}$

8. phi

but g(x) is also the messy $$\frac{-3x-7}{x-1}$$ so we put that expression in for g(x)

9. phi

btw, for future reference metal (iron or copper) is different from medal (shiny doodad)

10. phi

everywhere you see g(x) in $\frac{g(x) -7}{g(x)+3}$ erase the g(x) and replace it with $\frac{-3x-7}{x-1}$ you get a big mess, but it will simplify with some algebra

11. anonymous

(-3x-7)-7 --------- (x-1)+3

12. anonymous

?

13. phi

$\frac{\frac{-3x-7}{x-1}-7}{\frac{-3x-7}{x-1}+3}$

14. phi

that is how you replace g(x) with the messy fraction

15. anonymous

what happens next

16. phi

to simplify , I would multiply top and bottom by (x-1) $\frac{\left(\frac{-3x-7}{x-1}-7\right)}{\left(\frac{-3x-7}{x-1}+3\right)} \cdot \frac{(x-1)}{(x-1)}$

17. anonymous

ok so would that cancel out x-1 or...

18. phi

do the top first: distribute the (x-1) (which means multiply both terms inside by (x-1)

19. phi

the top is $\left(\frac{-3x-7}{x-1}-7\right)(x-1) \\ \frac{-3x-7}{(x-1)}(x-1) -7(x-1)$

20. phi

can you simplify the top ?

21. anonymous

yes, -3x ----- (x-1) x-1 I'm not sure

22. phi

$\frac{-3x-7}{(x-1)}(x-1)$ that is the first term, the (x-1) cancels

23. anonymous

if the bottom cancels then the top is -3x-7

24. phi

yes. and the 2nd term is -7(x-1) if you distribute the -7 what do you get ?

25. anonymous

-3x-7= -7x+7

26. phi

there is no = sign, just -3x-7+ -7x + 7 (that mess simplified to that) now combine like terms

27. anonymous

-10x

28. phi

yes the top (of the original fraction) simplifies to -10x now we do the bottom $\left(\frac{-3x-7}{x-1}+3\right)(x-1)$

29. anonymous

you said the bottom right?

30. phi

the (x-1) times the first term will cancel the (x-1) in the botom

31. phi

in case you lost track, we are doing the bottom of this mess: $\frac{\left(\frac{-3x-7}{x-1}-7\right)}{\left(\frac{-3x-7}{x-1}+3\right)} \cdot \frac{(x-1)}{(x-1)}$

32. phi

we found that $\frac{\left(\frac{-3x-7}{x-1}-7\right)}{\left(\frac{-3x-7}{x-1}+3\right)} \cdot \frac{(x-1)}{(x-1)} = \frac{-10x}{\left(\frac{-3x-7}{x-1}+3\right)(x-1)}$ and we are doing the bottom $\left(\frac{-3x-7}{x-1}+3\right)(x-1)$

33. phi

and the bottom $\left(\frac{-3x-7}{x-1}+3\right)(x-1)\\ \frac{-3x-7}{(x-1)}(x-1)+3(x-1)$

34. anonymous

x-1 cancels out so 3-7 is 4 so -3-4.

35. anonymous

-3x-4

36. phi

x-1 cancels so you are left with -3x-7 (for the first term)

37. phi

$\frac{-3x-7}{\cancel{(x-1)}}\cancel{(x-1)}+3(x-1) \\ -3x-7 +3(x-1)$

38. phi

now distribute the 3 in the 2nd term

39. anonymous

-3x-7+3x-3

40. anonymous

-10

41. phi

ok, so the bottom simplifies to -10 we found that $f(x) = \frac{\frac{-3x-7}{x-1}-7}{\frac{-3x-7}{x-1}+3} = \frac{-10x}{-10}$ one more step to go

42. phi

* f( g(x) ) =

43. anonymous

1?

44. phi

you can't ignore the x remember if you have $\frac{-10 \cdot x }{-10}$ that is the same as $\frac{-10}{-10} \cdot \frac{x}{1}$

45. anonymous

1x

46. phi

yes or just x you just showed f( g(x) ) = x the question was ***Confirm that f and g are inverses by showing that f(g(x)) = x and g(f(x)) = x. *** you just did the first part. now they want to do it the "other way" show g( f(x) ) = x I hope you are a "high-energy" person, because it's more work.

47. anonymous

is this right?

48. phi

it's almost right, g(x) is (-3x-7)/(x-1)

49. phi

it can get confusing unless we are slow and methodical $g(x) = \frac{-3x-7}{x-1} \\ g( f(x)) = \frac{-3f(x)-7}{f(x) -1}$

50. phi

now replace f(x) with its expression $g( f(x)) = \frac{-3\frac{x-7}{x+3}-7}{\frac{x-7}{x+3} -1}$

51. phi

follow?

52. anonymous

um not really

53. phi

do you follow this part $g(x) = \frac{-3x-7}{x-1} \\ g( f(x)) = \frac{-3f(x)-7}{f(x) -1}$

54. anonymous

yes

55. phi

and f(x) is "short-hand" (or the name for) $\frac{x-7}{x+3}$ so everywhere we see f(x), we can put in the "long form"

56. phi

$g( f(x)) = \frac{-3f(x)-7}{f(x) -1}$ the top says "multiply -3 times f(x)" then subtract 7 in math, and replacing f(x) with its expression, we would write -3 * (x-7)/(x+3) then subtract 7: -3 * (x-7)/(x+3) -7 $-3 \cdot \frac{x-7}{x+3} - 7$

57. anonymous

oh yea cancel out x-3 right

58. anonymous

x+3

59. phi

we haven't gotten that far, but that will be the idea. First, do you see how we get $g( f(x)) = \frac{-3\frac{x-7}{x+3}-7}{\frac{x-7}{x+3} -1}$

60. anonymous

yes I see that now.

61. phi

now you see the (x+3) that would be nice to get rid of. multiply top and bottom by (x+3)

62. anonymous

yes

63. phi

$\frac{\left(-3\cdot \frac{x-7}{x+3}-7\right)(x+3)}{\left(\frac{x-7}{x+3} -1\right)(x+3)}$

64. phi

can you do the top ? post your steps. the first step is "distribute" the (x+3)

65. anonymous

I thought x+3 cancels out?

66. phi

distribute means multiply each term inside the parens for example: (a + b)(x+3) = a(x+3) + b(x+3) use that same rule on $\left(-3\cdot \frac{x-7}{x+3}-7\right)(x+3)$

67. anonymous

-3(x-7)-7 -3x+21-7 -3x+14 ?

68. phi

"terms" are things multiplied together. the -3 * (x-7)/(x+3) is all one term terms are separated by + or -. in other words, the -7 is also a term Here is what it looks like, after distributing the (x+3) $\left(-3\cdot \frac{x-7}{x+3}-7\right)(x+3)\\ -3\cdot \frac{x-7}{x+3}\cdot (x+3) -7 \cdot (x+3)$ notice the -7 is also multiplied by (x+3)

69. phi

you did the first part correctly. -3(x-7) but you also have -7(x+3)

70. anonymous

-3x+14-7x-21 4x-7

71. anonymous

-10x-7

72. phi

-3(x-7) + -7(x+3) try again

73. anonymous

-3x-7+-7x+21 -10x+14

74. phi

ok but how are you doing -3(x-7) ?

75. anonymous

ugh I messed up again. -3x+21+-7x-21

76. anonymous

-10x

77. phi

yes, that looks good. now we do the bottom $\left(\frac{x-7}{x+3} -1\right)(x+3)$

78. phi

any luck?

79. anonymous

no luck :(

80. phi

can you distribute (x+3) by writing (x+3) next to each term inside the parens?

81. anonymous

(x-7) -1(x+3)

82. phi

ok, now distribute the -1

83. anonymous

x-7-1x-3

84. anonymous

-10

85. phi

ok, so we have $g( f(x) ) = \frac{-10x}{-10}$

86. anonymous

x

87. phi

yes, so you have shown that f(g(x))= x and g(f(x))= x which proves that f(x) and g(x) are inverses.

88. anonymous

Thank you soo much :) can we do one more problem it won't take that long.

89. phi

if you go fast

90. anonymous

Find the angle θ (if it exists) in the interval [0°, 90°) for which sin θ = cos θ. θ = 30° θ = 45° No such angle exists. θ = 60°

91. anonymous

options^

92. phi

divide both sides by cos theta

93. anonymous

45 degs

94. phi

yes, you get sin x/cos x = 1 tan x = 1 x= atan 1 = 45º

95. phi

or you can remember that sin 45 = sqr(2)/2 and cos 45 = the same thing