Question

--

Answer 1

what is your question?

Answer 2

we have 1/x

x = 0

so replace x with 0, since x = 0, to get 1/x = 1/0

what do we get when we try to evaluate 1/0 ?

Answer 3

yes because you cannot divide by zero

Answer 4

so what kind of discontinuity?

Answer 5

good

Answer 6

what do you mean?

Answer 7

correct, for the same reason (you cannot divide by zero)

Answer 8

sure go for it

Answer 9

for the 1+ kx , its x greater than or equal to 5

Answer 10

for each piece, what happens when x = 5?

Answer 11

for the first it turns into 15 - k and the second 1+5k

Answer 12

so if we want this function to be continuous at x = 5, this must mean that the two pieces produce the same y output when x = 5

Answer 13

so essentially, they must be equal to each other when x = 5, so

15 - k = 1+5k

now you solve for k

Answer 14

@jim_thompson5910

Answer 15

I'm getting the same answer, nice job

Answer 16

great!!

Answer 17

first factor the numerator and denominator

Answer 18

hmm now that I look at it, this is a cubic which is a pain to factor

also, the cubic is x^3 - 3x^2 - 2x + 1 right?

Answer 19

ok got it

Answer 20

I don't think x^3 - 3x^2 - 2x + 1 factors at all

Answer 21

is there a typo in x^3 - 3x^2 - 2x + 1 ?

Answer 22

oh is it x^3 - 3x^2 - 2x + 6 ?

Answer 23

ok I get (x^2 - 2)(x-3) now

Answer 24

the x-3 terms cancel

Answer 25

what discontinuity is associated with what we just canceled out?

Answer 26

no, x = 3 is removable because we have that cancellation

Answer 27

when you cancel out x-3, you are left with (x^2 - 2)/(x+3)

Answer 28

x = 3 is allowed in the domain of (x^2 - 2)/(x+3)

so there are no discontinuities at x = 3 in (x^2 - 2)/(x+3)

Answer 29

however, to make sure the original expression (x^3 - 3x^2 - 2x + 6)/(x^2-9) and the expression (x^2 - 2)/(x+3) are equivalent, we have to remove x = 3 from the domain

this is because x = 3 causes problems for the first expression

Answer 30

it is a removable discontinuity based on the reasoning I gave above

Answer 31

I recommend you compare the graphs of (x^3 - 3x^2 - 2x + 6)/(x^2-9) and (x^2 - 2)/(x+3)

Answer 32

one sec

Answer 33

I'm graphing using geogebra and I'll attach it

Answer 34

actually that wont work, nvm

Answer 35

notice how in the first expression (x^3 - 3x^2 - 2x + 6)/(x^2-9), if we plug in x = 3, we get 0 in the denominator

however, if we plug that same value x = 3 into (x^2 - 2)/(x+3) we don't get 0 in the denominator

Answer 36

so x = 3 causes problems for (x^3 - 3x^2 - 2x + 6)/(x^2-9) but not for (x^2 - 2)/(x+3)

Answer 37

the two expressions (x^3 - 3x^2 - 2x + 6)/(x^2-9) and (x^2 - 2)/(x+3) are effectively equal

they produce the same outputs for every x value in their domains...it's just that the first expression has a slightly different domain: it excludes x = 3 while the second does not exclude x = 3

Answer 38

to ensure the two expressions are fully equivalent, for all input x values, we must exclude x = 3 from both expressions

so that is why x = 3 is a removable discontinuity

Answer 39

hopefully this makes sense?

Answer 40

no, by default if you don't mention "exclude or kick out x = 3" the second expression (x^2 - 2)/(x+3) would happily allow x = 3 into the domain

x = 3 works just fine: it produces some y value when you plug it in

Answer 41

so you have to add in those instructions to ensure the two expressions are equal for all allowed x values

Answer 42

I'm not sure what you're asking to be honest

Answer 43

are you asking about the steps to simplify?

Answer 44

that's ok

Answer 45

and yes, you factor and then cancel out the common terms to simplify

Answer 46

well the first step is to find all of the discontinuities of the original expression

Answer 47

what are those discontinuities?

Answer 48

the original denominator is x^2 - 9

Answer 49

set that equal to zero and solve for x

Answer 50

don't worry about the cancellation yet

Answer 51

just focus on solving x^2 - 9 = 0

Answer 52

doing that gives x = -3 or x = 3

Answer 53

now when you cancel out the x-3 terms, only x+3 is left

x+3 means x = -3 when you set it equal to zero

so notice how the discontinuity x = 3 was removed

Answer 54

technically it's still a discontinuity

Answer 55

hmm not sure what else to add

Answer 56

but you can think of them as a point you remove from the graph (a single point)

contrast that with a vertical asymptote

Answer 57

I can help when you get back