4) Find all values where the function is discontinuous.
f(x) = 4x/((3x-1)(2x+5))

- anonymous

4) Find all values where the function is discontinuous.
f(x) = 4x/((3x-1)(2x+5))

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- schrodinger

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- anonymous

Btw this is just review for my midterm on my week points for calculus.

- amistre64

there are relatively few cardinal sins in math; the biggest is dividing by a zero

- anonymous

So i really appreciate the help, I have been so busy with my senior project and my other class that I have had very little time to study.

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## More answers

- amistre64

what values of x make the bottom go zero?

- anonymous

Finding the zero's?

- amistre64

the zeros of the bottom, not the zeros of the function perse

- anonymous

I remember trying to solve for zero

- amistre64

then solve for zeros of the bottom:
(3x-1)(2x+5) = 0, for what values of x?

- anonymous

making the equation equal to zero?

- anonymous

yes exactly

- amistre64

recall your multiplication tables .... 0 was one of the simplest ones to remember

- amistre64

given a product of a and b
ab = 0 when a=0, or b=0, or both equal 0

- anonymous

Anything times zero is zero lol

- amistre64

yes :)
so
(3x-1)(2x+5) = 0, when 3x-1 = 0 , or when 2x+5 = 0

- anonymous

4x/((3x-1)(2x+5))
4x = 0
3x -1 = 0
2x + 5 = 0

- amistre64

4x=0 is not a cardinal sin in this case. we are totally allowed to have a zero on top of a fraction

- amistre64

a function is discontinuous where it is undefined, or where it breaks apart

- amistre64

a fraction is undefined when it has a zero underneath:
n/0 is bad

- anonymous

I get that

- amistre64

so we see that this function is undefined when:
3x -1 = 0
or
2x + 5 = 0
so all we have to do is determine for what values of x that happens

- anonymous

So the numerator is allow to have a zero but the denominator is not allowed

- amistre64

correct

- anonymous

So fill in values for x?

- amistre64

of course, or algebra out some values

- amistre64

im not real sure what process youre thinking of there

- amistre64

3x - 1 = 0 , to solve for x, lets start by adding 1 to each side
+ 1 +1
----------
3x + 0 = 1 , since anything +0 is itself (indentity), this simplifies to
3x = 1 , now to get rid of the 3 stuck there, we divide it off, 3/3 = 1
/3 /3
-------
1x = 1/3 , since anything times 1 is itself (identity), this simplifies to
x = 1/3
the same concepts can be applied to the other factor

- anonymous

3x -1 = 0
3x = 0 + 1
or
2x + 5 = 0
2x = 0 + 5

- amistre64

2x = -5 , but yeah

- anonymous

yeah i forgot to flip the sign

- amistre64

if your intent is to "flip the sign" and "move things to the other side"; then you are not applying any sound mathematical principals to this

- amistre64

there are 5 simple properties to algebra, and they can prevent a world of mistakes

- amistre64

1) ab = ba , commutative property
2) (ab)c = a(bc) , associative property
3) \(aa^{-1}=e\), inverse property; the inverse in addition is the negative, the inverse in multiplication is dividing by the reciprocal
4) \(ae=a\), identity property; the identity in addition is +0, the identity in multiplication is *1
5) a(b+c) =ab + ac , distributive property ... multiplication distributes over addition

- amistre64

2x + 5 = 0 , addition inverse, -5
- 5 -5
-----------
2x + 0 = -5 , addition identity
2x = -5 , multiplicative inverse, /2
/2 /2
--------
1x = -5/2 , multiplicative identity
x = -5/2 , multiplicative identity

- anonymous

okay so then we try to get x alone

- anonymous

I understand that

- amistre64

yes, by using the properties of algebra, we can isolate the variable

- anonymous

thanks

- anonymous

signs reverse on the other side of the equation

- amistre64

:) or make up some properties as we go .... the sign reverse on the other side property for example.
head math is great, but it has its drawbacks is all

- anonymous

okay so the inverse becomes the identity

- amistre64

correct, applying the inverse of a given element, produces the identity element

- anonymous

inverse meaning the opposite and identity the after math

- amistre64

yep

- anonymous

so this function is discontinuous at x = -5/2

- amistre64

that is one of the places yes

- amistre64

the other one is at x=1/3

- amistre64

the strategy in determining discontinuity is in finding the parts of the function that can "go bad"
dividing by 0 is bad
taking the even-root of a negative number is bad among the Reals
taking the logarothm of 0 or less is bad
those seem to be the big ones to watch for

- anonymous

Okay thanks you up for another one with lim?

- amistre64

i can take a stab at it .. i only have so much wits about me in a day before I start to go stupid lol

- anonymous

lmao yeah thats how I feel right about now

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