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Janu16

  • one year ago

Given the function f(x)=4(x+3)-5, solve for the inverse function when x=3.

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  1. Janu16
    • one year ago
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    plz help

  2. Janu16
    • one year ago
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    @Mehek14

  3. Janu16
    • one year ago
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    @Michele_Laino

  4. anonymous
    • one year ago
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    To find the inverse of a function, you always want to change x and y, then solve for y again. First, change f(x) to y to avoid any confusion. \[y=4(x+3)-5\] Now, swap x and y. \[x=4(y+3)-5\] All you have to do now is solve for y. Once you've done that, substitute 3 in for x, and you'll be all set!

  5. Janu16
    • one year ago
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    Wait so how do you substitute x for 3?

  6. Janu16
    • one year ago
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    @adrianna45

  7. Janu16
    • one year ago
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    @nincompoop can you please help?

  8. Janu16
    • one year ago
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    @perl

  9. geerky42
    • one year ago
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    by plug in x=3

  10. geerky42
    • one year ago
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    Just replace x to 3.

  11. Janu16
    • one year ago
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    so it would be x=4(y+3)-5 x=4(3+3)-5?

  12. Janu16
    • one year ago
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    @geerky42

  13. DanJS
    • one year ago
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    Are you still working on this ?

  14. Janu16
    • one year ago
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    yes

  15. DanJS
    • one year ago
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    f(x) = y = 4(x+3) - 5 is a line that is given y = 4x + 12 -5 y = 4x + 7

  16. Janu16
    • one year ago
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    ya so when do you substitute x for 3?

  17. DanJS
    • one year ago
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    The inverse of that , will be another line that is reflected over the line with slope 1, y=x. To figure that , you can just switch x and y around and resolve for the new y.

  18. DanJS
    • one year ago
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    Original: y = 4x + 7 Inverse: x = 4y + 7 x - 7 = 4y y = (x-7) / 4

  19. DanJS
    • one year ago
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    Put x=3 in the inverse function, calculate the y value

  20. Janu16
    • one year ago
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    -1?

  21. geerky42
    • one year ago
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    That's correct.

  22. DanJS
    • one year ago
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  23. DanJS
    • one year ago
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    See it is a reflection if you switch the x and y around, as in the graph

  24. Janu16
    • one year ago
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    ya i see it. it really helped thank you so much and do you know how to do f(g(x) problems?

  25. DanJS
    • one year ago
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    sure

  26. Janu16
    • one year ago
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    well i need help with Given that f(x) = x2 + 2x + 3 and g(x) = quantity of x plus four, over three, solve for f(g(x)) when x = 2.

  27. Janu16
    • one year ago
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    like i dont know how to setup

  28. DanJS
    • one year ago
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    solve g(2) first, what is that

  29. DanJS
    • one year ago
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    g(x) = (x+4) / 3

  30. Janu16
    • one year ago
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    2?

  31. Janu16
    • one year ago
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    wait do you put 2 for x?

  32. DanJS
    • one year ago
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    yes, now find f[g(2)], what do ya think to do?

  33. Janu16
    • one year ago
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    4x2x+3 than 6x=3 which is 2?

  34. DanJS
    • one year ago
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    since g(2) = 2 , start from the middle and work outwards f(2) = 2^2 + 2*2 + 3

  35. Janu16
    • one year ago
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    11?

  36. DanJS
    • one year ago
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    yeah

  37. Janu16
    • one year ago
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    oh i forgot to substitute x with 2 for 2x

  38. Janu16
    • one year ago
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    so if i haveanother problem like in future do i solve g first? or depending on the problem

  39. DanJS
    • one year ago
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    for compound functions like f([g(x)] start with the inner function, evaluate if given a value, then use g(x) value in f(x)... work inner function to outer functions

  40. Janu16
    • one year ago
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    ohok thank you very much!

  41. DanJS
    • one year ago
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    welcome, goodluck

  42. Janu16
    • one year ago
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    there are like some problems like f(x) and g(x) and you have to compare them. there are lot of problems but i dont know how to compare like if you know how to do it i will put it on new page

  43. DanJS
    • one year ago
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    sure, just tag my name with @name

  44. Janu16
    • one year ago
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    ok thanks

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