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Destinyyyy

  • one year ago

Help with difference quotient..

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  1. Destinyyyy
    • one year ago
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  2. Destinyyyy
    • one year ago
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    @Nnesha

  3. Nnesha
    • one year ago
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    substitute x for x+h

  4. Destinyyyy
    • one year ago
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    So... f(x)-f(x) / h

  5. Nnesha
    • one year ago
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    function is f(x) = sqrt{14x} <--replace x with x+h to find f(x+h)

  6. Nnesha
    • one year ago
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    formula is \[\large\rm \frac{ f(\color{ReD}{x+h}) -f(x) }{ h }\]

  7. Destinyyyy
    • one year ago
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    So square root 14(x+h)

  8. Destinyyyy
    • one year ago
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    Im at.. square root 14 (x+h-x) over h(square root x+h + square root x)

  9. Nnesha
    • one year ago
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    yes right so \[\huge\rm \frac{ \sqrt{14(x+h)} - \sqrt{14x} }{ h }\]

  10. Nnesha
    • one year ago
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    eh eh what nope you can't combine both sqrts

  11. Destinyyyy
    • one year ago
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    ?

  12. Nnesha
    • one year ago
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    hmm how did you get square root x+h + square root x) at the denominator ?

  13. Destinyyyy
    • one year ago
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    Final answer--> square root 14 over square root x+h + square root x

  14. Nnesha
    • one year ago
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    sqrt{14(x+h) } can be written as sqrt{14} sqrt{x+h) so \[\frac{ \sqrt{14}\sqrt{x+h} -\sqrt{14}\sqrt{x}}{ h }\] sqrt 14 is common so you can take it out

  15. Destinyyyy
    • one year ago
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    Um not sure where your at

  16. Nnesha
    • one year ago
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    \[\huge\rm \frac{ \sqrt{14} }{ \sqrt{x+h}+\sqrt{x}}\] is this ur final answer ?

  17. Destinyyyy
    • one year ago
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    Yes and its correct.

  18. Nnesha
    • one year ago
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    ahh i see so you have to multiply top and bottom with the conjugate of the numerator

  19. Destinyyyy
    • one year ago
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    Yup

  20. Nnesha
    • one year ago
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    sqrt{14(x+h) } can be written as sqrt{14} sqrt{x+h) so \[\huge\rm \frac{ \sqrt{14}\sqrt{x+h} -\sqrt{14}\sqrt{x}}{ h }\] sqrt 14 is common so you can take it out did you understand that step ?

  21. Destinyyyy
    • one year ago
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    Yeah it gets moved to the front --- square root 14 ( square root x+h) - square root x over h ... That was the second step

  22. Destinyyyy
    • one year ago
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    Thanks for helping.

  23. Nnesha
    • one year ago
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    yw.

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