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kymber Group Title

Composition of functions?

  • 2 years ago
  • 2 years ago

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  1. kymber Group Title
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    • 2 years ago
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  2. ParthKohli Group Title
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    Yes ma'am.

    • 2 years ago
  3. ParthKohli Group Title
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    I think that this is pretty easy for you. We know,\[(g \circ f)(x) = g(f(x)) \]Therefore, our composition simplifies as the following:\[(g \circ f)(x) \implies g(f(x)) \implies g(2x - 1) \implies (2x - 1) +2 \]

    • 2 years ago
  4. ParthKohli Group Title
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    Do you get the above?

    • 2 years ago
  5. ParthKohli Group Title
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    ... or should I be more informal? lol

    • 2 years ago
  6. kymber Group Title
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    I don't understand what composition means :I

    • 2 years ago
  7. ParthKohli Group Title
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    Never mind that "composition" word. lol

    • 2 years ago
  8. ParthKohli Group Title
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    Do you still understand my explanation?

    • 2 years ago
  9. kymber Group Title
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    No. I don't understand what I'm supposed to be doing at all! :'(

    • 2 years ago
  10. ParthKohli Group Title
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    I'd give you an example.\[\sqrt{x + 1} \]We're doing two things to \(x\). First, we are finding the square root of x, then we are adding 1 to x. So,\[x + 1\]is the composition of two functions: \(\sqrt{\text{stuff}}\) and \(\text{stuff} + 1\).

    • 2 years ago
  11. ParthKohli Group Title
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    Here, we are again doing two things to \(x\). First, we are multiplying two to it and adding 1(which is \(2x - 1\)) Then, we are adding two to that number.

    • 2 years ago
  12. ParthKohli Group Title
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    Operation Multiply Two And Subtract 1 got a boring name, which is f(x). Operation Add Two To Whatever Comes In Our Way got the name g(x).

    • 2 years ago
  13. ParthKohli Group Title
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    \(x\) went to Operation Multiply Two And Subtract 1, and then became \(2x - 1\). Sad :(

    • 2 years ago
  14. ParthKohli Group Title
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    And then it went to Operation Add Two To Whatever Comes In Our Way and became \(2x - 1 + 2 = 2x + 1\)

    • 2 years ago
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