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anonymous

  • one year ago

For the function, find and simplify...will write equation below

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  1. anonymous
    • one year ago
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    (A) \[f(x+h)-f(x)\] and (B)\[\frac{ f(h+h)-f(x) }{ h }\] \[f(x)=9x^2-1\]

  2. butterflydreamer
    • one year ago
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    We know our function is: f(x)=9x^2−1 For part (a), it wants you to find f( x + h) - f(x), So firstly you need to work out what "f (x+h)" equals to... To do this, you just sub (or "plug") in "x+h" into your original f(x) function... so replace the x's with (x+h)

  3. butterflydreamer
    • one year ago
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    |dw:1434345522113:dw|

  4. UsukiDoll
    • one year ago
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    isn't it this function for b which is also the definition of the derivative in calculus I? \[\frac{ f(x+h)-f(x) }{ h }\]

  5. anonymous
    • one year ago
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    I honestly have no clue how to do this so I'm not sure

  6. butterflydreamer
    • one year ago
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    are you confused on how to find f(x+h) ?

  7. anonymous
    • one year ago
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    So I only have one example and here's how it goes...\[F(x)=x^2+3x\] Step 1: \[(x+h)^2+3(x+h)\] Now with this example that I opened the question with...I dont know how to set it up like this example

  8. butterflydreamer
    • one year ago
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    In that example, your function is f(x) = x^2 + 3x They found f(x + h) by REPLACING the "x" in "f(x) = x^2 + 3x" with "x+h" That's why we get: f(x + h) = (x + h) ^2 + 3 (x + h)

  9. butterflydreamer
    • one year ago
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    |dw:1434346176107:dw|

  10. anonymous
    • one year ago
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    So I think I'm confused because I don't know where the 9 should go because in the example I just showed you, x was 1

  11. butterflydreamer
    • one year ago
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    9 stays the same... Your entire function will remain exactly the same but you just replace the "x" with "x+h"

  12. anonymous
    • one year ago
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    So it should be (9x+h)^2-1(x+h)

  13. butterflydreamer
    • one year ago
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    no.. we ONLY replace the X. ORIGINAL function f(x) = 9x^2 -1 f(x+h) = 9(x+h)^2 - 1 we only replaced x with x+h We kept the 9 and the 1.

  14. anonymous
    • one year ago
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    oh ok. I think I can handle the rest now. I just didn't know how to set up my problem

  15. butterflydreamer
    • one year ago
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    sureee, good luck !

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