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## math456 3 years ago Alright last question need help! Q.51

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1. math456

2. math456

@lgbasallote @jim_thompson5910

3. nphuongsun93

to me, the a and b are discontinuous at a, because they are undefined when x = a

4. math456

right so when I plug in 1 is get -1 and when i plug 2 i get 15.. so they aren't equal!

5. nphuongsun93

lol no a) if you plug in x = 1, the denominator will be 1-1 = 0 --> undefined b) if you plug in x = 2, the denominator will be 2-2 = 0 --> undefined now you have to find function g (try simplify the function f for a and b)

6. nphuongsun93

i will do a) for you $f(x) = \frac{x^4-1}{x-1}$ using a^2 - b^2 = (a+b)(a-b) $f(x) = \frac{(x^2+1)(x^2-1)}{x-1}$and again$f(x) = \frac{(x^2+1)(x+1)(x-1)}{x-1}$simplify$f(x) = (x^2+1)(x+1)$^^^ your new function g

7. math456

but its hard. should i find it by the cal..

8. math456

ohhh i was asking for Question 51 not 47!

9. math456

the IVT theorem

10. nphuongsun93

Oh sheet my bad

11. nphuongsun93

x^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 x-axis somewhere, therefore there's a root between 1 and 2.

12. math456

so I was right b4 :P

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