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anonymous

  • 5 years ago

difference quotient of function f(x) is defined as : f(x+h) - f(x)/ h compute the difference quotient of the given functions a) y(z) = 1/ z+2 b) g(x)= 6-x^2 please help

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  1. anonymous
    • 5 years ago
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    the h is the denominator

  2. anonymous
    • 5 years ago
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    I think it should be [f(x+h) - f(x)/ ]h instead of f(x+h) - f(x)/ h

  3. anonymous
    • 5 years ago
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    I am sorry for the typos I meant "I think it should be [f(x+h) - f(x) ]/h instead of f(x+h) - f(x)/ h"

  4. anonymous
    • 5 years ago
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    yeah iamignorant

  5. anonymous
    • 5 years ago
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    y(z +h) -y(z)= 1/(z+h+2) - 1/(z+2)= z+2-z-h-2/(z+h+2)(z+2) = -h/(z+h+2)(z+2) dividing by h on both sides gives y(z +h) -y(z)/h = 1/(z+h+2)(z+2)

  6. anonymous
    • 5 years ago
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    oh i missed the negative sign in the numerator :)

  7. anonymous
    • 5 years ago
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    g(x) = 6 - x^2 g(x+h) = 6-(x +h)^2= 6-x^2-2xh- h^2 g(x+h) - g(x) = -2xh - h^2 = -h(2x +h) g(x+h) - g(x)/h = 2x +h

  8. anonymous
    • 5 years ago
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    do u mind explaining the y(z+h)-y(z) blablabla how it comes about?

  9. anonymous
    • 5 years ago
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    replace z by z +h

  10. anonymous
    • 5 years ago
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    then simlify the expression by taking the LCM

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