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anonymous

  • one year ago

Can you help me understand how to do this problem. Given the function f(x) = 4(x+3) - 5, solve for the inverse function when x = 3. (1 point)

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  1. anonymous
    • one year ago
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    To make this easier, let y = f(x). So, we have: \[y = 4(x + 3) - 5\]To solve for the inverse, switch x and y. \[x = 4(y + 3) - 5\]Now, solve for y again in terms for x. Can you take it from there?

  2. anonymous
    • one year ago
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    3=4y+12 - 5 so 3=4y+7 so -4=4y so y=-1?

  3. anonymous
    • one year ago
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    Yes! Good job! Also, there's a cool little trick to inverse functions like this. If you're trying to find what the inverse is at an x-value, you can just plug that in for y (or f(x)) in the original and solve for x.

  4. anonymous
    • one year ago
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    Can you give me an example? Im not sure I understand that.

  5. anonymous
    • one year ago
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    what about this when you don't have a value for x and need to do the inverse? Given the function f(x) = 5x-3/4 , which of the below expressions is correct? f-1(x) = 3-5x/4 f-1(x) = 4x+3/5 f-1(x) = 4x-3/5 f-1(x) = -5x-3/4

  6. anonymous
    • one year ago
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    Follow the same steps as in the first example, but this time, since you don't have any values to plug in, leave the x there and solve for y in terms of x. Then, once you get y by itself, replace it with \(f^{-1}(x)\).

  7. anonymous
    • one year ago
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    So, to get you started, \[f(x) = 5x-\frac34\]\[y=5x-\frac34\]\[x=5y-\frac34\]

  8. anonymous
    • one year ago
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    it's actually 5x-3 over 4

  9. anonymous
    • one year ago
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    it wouldn't copy the equation correctly.

  10. anonymous
    • one year ago
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    Ah ok. No problem. \[f(x) = \frac{5x-3}{4}\]\[y=\frac{5x-3}{4}\]\[x=\frac{5y-3}{4}\]

  11. anonymous
    • one year ago
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    then how do i do the f-1(x)

  12. UsukiDoll
    • one year ago
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    split the fractions up \[x = \frac{5y}{4}- \frac{3}{4}\]

  13. UsukiDoll
    • one year ago
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    now add 3/4 to both sides.

  14. anonymous
    • one year ago
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    Well, remember when we let y = f(x)? Solve for y first, then you can just replace the y with the \(f^{-1}(x)\). It's just a notation thing. You can work with f(x) the entire time, it's just a bit messy.

  15. anonymous
    • one year ago
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    Do you need help with the algebra to solve for y or can you do that part?

  16. anonymous
    • one year ago
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    can you help? I got y=3+4X? Is that right?

  17. UsukiDoll
    • one year ago
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    let me check that for a sec

  18. anonymous
    • one year ago
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    Not quite. I think you forgot the 5 somewhere.

  19. UsukiDoll
    • one year ago
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    \[x + \frac{3}{4}= \frac{5y}{4}\] then divide 5/4 throughout the entire equation .

  20. anonymous
    • one year ago
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    \[x = \frac{5y - 3}{4}\]\[4x = 5y - 3\]\[5y = 4x + 3\]It looks like you got to here in your algebra. All that's left to do is divide by 5 to get y by itself. Do you get it?

  21. UsukiDoll
    • one year ago
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    |dw:1439784708621:dw||dw:1439784747500:dw|

  22. anonymous
    • one year ago
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    I multiplied the entire equation by 4 in the first step, then added 3 on both sides in those first two steps.

  23. anonymous
    • one year ago
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    okay, i get that!

  24. UsukiDoll
    • one year ago
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    |dw:1439784792258:dw| it was a good attempt. just that the 5 was missing ^_^

  25. anonymous
    • one year ago
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    Remember, you're basically doing the opposite of PEMDAS to solve for a variable. Just work backwards!

  26. UsukiDoll
    • one year ago
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    now the last step is to replace that y with \[f^{-1}(x )\]

  27. anonymous
    • one year ago
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    okay thank you!

  28. anonymous
    • one year ago
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    Yes ^. Like I said before, we let y = f(x) to make it more convenient when writing (It'd take a while to write f(x) and f^-1(x) every time!). So now, you can just replace y with \(f^{-1}(x)\). You can decide to solve it using the f(x), and it'd still be correct, it'd just take longer to write and is a bit more confusing if you're new to algebra. :)

  29. UsukiDoll
    • one year ago
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    |dw:1439785031869:dw|

  30. anonymous
    • one year ago
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    Do you know how to explain how to do this? On the graph of the equation 3x + 2y = 18, what is the value of the y-intercept?

  31. anonymous
    • one year ago
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    I'm going to sleep in a few minutes, but this last one, sure. A y-intercept is basically where the graph touches the y-axis. What do we know about the x-value at the y-axis?

  32. anonymous
    • one year ago
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    That's the only information it gives me

  33. anonymous
    • one year ago
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    and these are the answer choices -9 -6 6 9

  34. UsukiDoll
    • one year ago
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    for this equation 3x+2y=18 we need this in y = mx+b form where m is the slope and b i s the y -intercept. You need to rearrange this equation and divide the whole equation by 2

  35. anonymous
    • one year ago
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    I know, I'm trying to get you through the question without my outright giving you how to do it.

  36. anonymous
    • one year ago
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    @UsukiDoll I personally think there's a much easier way to do this question, but that works too.

  37. anonymous
    • one year ago
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    ah okay. Well i put it in y=mx+b format, then I wasn't sure what to do

  38. anonymous
    • one year ago
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    If you did put it into y = mx + b, then b is your y-intercept by definition.

  39. anonymous
    • one year ago
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    So the answer would be 9

  40. anonymous
    • one year ago
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    Yes, it is 9. The point I was trying to make earlier is this. You learned in algebra that the y-intercept is where the graph hits the y-axis, correct?

  41. anonymous
    • one year ago
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    Yes that makes sense! Thank you for your help! I appreciate it!

  42. UsukiDoll
    • one year ago
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    2y=-3x+18 y = -3/2x+9 -3/2 <- slope 9<-y-intercept.

  43. anonymous
    • one year ago
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    At the y-axis, the x-value of all points on the y-axis is 0. In other words, the y-intercept is when x = 0. So, another way to do it (personally faster) is to just let x = 0 and solve for y. \[3x + 2y = 18\]\[3(0) + 2y = 18\]\[2y = 18\]\[y = 9\]

  44. anonymous
    • one year ago
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    However, either way works, of course. :)

  45. UsukiDoll
    • one year ago
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    oh .. that's what you mean by easier way. I never thought of that when I used to to do these problems. for y-intercept, set x = 0 and solve for y XD

  46. UsukiDoll
    • one year ago
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    I was taught to rearrange to y = mx+b and if there's a number next to y, divide the whole equation by that number. wow tsk tsk. the math professors keeping their good secrets lol

  47. anonymous
    • one year ago
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    y=mx+b is how I learned, but that makes sense.

  48. anonymous
    • one year ago
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    lol yeah, lots of ways to think about these types of problems. It's just in the beginning, it's hard to grasp the concept, so you learn rules and directions like (just put into y = mx + b first). However, as you get more used to it, you can start to think about it more and you get little tricks like that.

  49. anonymous
    • one year ago
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    Anyway, I'm going to sleep. I'm sure if you have anymore questions, others will be glad to help you @gmanmoney .

  50. anonymous
    • one year ago
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    thanks so much!

  51. anonymous
    • one year ago
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    Brittany was tracking the increasing temperature in the morning. At 8 a.m. it was 68 degrees Fahrenheit. At 10 a.m. it was 74 degrees Fahrenheit. If Brittany made the function f(x) = 3x + 36, what would the 36 represent? (1 point) The temperature at midnight The rate the temperature was increasing The length of time she recorded for The total change in degrees

  52. anonymous
    • one year ago
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    How would I do this one?

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