Question

Alright last question need help! Q.51

Answer 1

@lgbasallote @jim_thompson5910

Answer 2

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

Answer 3

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

Answer 4

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)

Answer 5

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

Answer 6

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

Answer 7

ohhh i was asking for Question 51 not 47!

Answer 8

the IVT theorem

Answer 9

Oh sheet my bad

Answer 10

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.

Answer 11

so I was right b4 :P