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math456
 2 years ago
Best ResponseYou've already chosen the best response.1@lgbasallote @jim_thompson5910

nphuongsun93
 2 years ago
Best ResponseYou've already chosen the best response.1to me, the a and b are discontinuous at a, because they are undefined when x = a

math456
 2 years ago
Best ResponseYou've already chosen the best response.1right so when I plug in 1 is get 1 and when i plug 2 i get 15.. so they aren't equal!

nphuongsun93
 2 years ago
Best ResponseYou've already chosen the best response.1lol no a) if you plug in x = 1, the denominator will be 11 = 0 > undefined b) if you plug in x = 2, the denominator will be 22 = 0 > undefined now you have to find function g (try simplify the function f for a and b)

nphuongsun93
 2 years ago
Best ResponseYou've already chosen the best response.1i will do a) for you \[f(x) = \frac{x^41}{x1}\] using a^2  b^2 = (a+b)(ab) \[f(x) = \frac{(x^2+1)(x^21)}{x1}\]and again\[f(x) = \frac{(x^2+1)(x+1)(x1)}{x1}\]simplify\[f(x) = (x^2+1)(x+1)\]^^^ your new function g

math456
 2 years ago
Best ResponseYou've already chosen the best response.1but its hard. should i find it by the cal..

math456
 2 years ago
Best ResponseYou've already chosen the best response.1ohhh i was asking for Question 51 not 47!

nphuongsun93
 2 years ago
Best ResponseYou've already chosen the best response.1x^4 + x  3 = 0 1^4 + 1  3 = 2 2^4 + 2  3 = 15 to get from 2 to 15, you have to cross the xaxis somewhere, therefore there's a root between 1 and 2.
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