Difference Quotient Part 2

Discussion in 'Algebra' started by nycmathguy, Jul 4, 2021.

  1. nycmathguy

    nycmathguy

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    I want to continue with this question posted on 7/3/21.

    Find the difference quotient
    and simplify your answer.

    f(x) = x^(2/3) + 1

    [f(x) - f(8)]/(x - 8), where x cannot be 8.

    I will find f(8) first.

    Note: x^(2/3) = cuberoot{x^2}.

    cuberoot{x^2} + 1

    cuberoot{8^2} + 1

    cuberoot{64} + 1

    16 + 1 = 17

    [cuberoot{x^2} - 17]/( x - 8)

    Rationalize the numerator.

    Numerator:

    [cuberoot{x^2} - 17][cuberoot{x^2} + 17]

    Denominator:

    (x - 8)[cuberoot{x^2} + 17]

    Let cr = cube root

    [cr{x^2} - 17][cr{x^2} + 17]/(x - 8)[cr{x^2} + 17]

    Simplify the numerator. Leave denominator alone.

    Note: (cr{x^2})^2 = x^(4/3)

    Note: (-17)(17) = -289

    Finally, I came up with this:

    [x^(4/3) - 289]/(x - 8)

    I don't think this can be simplified further.

    You say?
     
    nycmathguy, Jul 4, 2021
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  2. nycmathguy

    MathLover1

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    you are using wrong formula

    f(x) = x^(2/3) + 1

    [f(x+h) - f(x)]/ h, where h cannot be 0

    first find f(x+h)

    f(x+h) = (x+h)^(2/3) + 1
    f(x+h) = (x+h)^(2/3) + 1

    then substitute in [f(x+h) - f(x)]/ h both f(x) and f(x+h)

    [ (x+h)^(2/3) + 1 - x^(2/3) + 1]/ h

    [ (h + x)^(2/3) - x^(2/3) + 2]/ h

    or
    (h + x)^(2/3) /h- x^(2/3)/h+ 2/ h
     
    MathLover1, Jul 4, 2021
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    nycmathguy likes this.
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  3. nycmathguy

    nycmathguy

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    Understood. Other than that, seriously, not too bad. Can you copy and paste your reply here to the Difference Quotient Part 1 thread?
     
    nycmathguy, Jul 4, 2021
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