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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)= 6x^2
please help
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)= 6x^2 please help

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the h is the denominator

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think it should be [f(x+h)  f(x)/ ]h instead of f(x+h)  f(x)/ h

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0y(z +h) y(z)= 1/(z+h+2)  1/(z+2)= z+2zh2/(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)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh i missed the negative sign in the numerator :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0g(x) = 6  x^2 g(x+h) = 6(x +h)^2= 6x^22xh h^2 g(x+h)  g(x) = 2xh  h^2 = h(2x +h) g(x+h)  g(x)/h = 2x +h

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do u mind explaining the y(z+h)y(z) blablabla how it comes about?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then simlify the expression by taking the LCM
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