anonymous
  • anonymous
4) Find all values where the function is discontinuous. f(x) = 4x/((3x-1)(2x+5))
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Btw this is just review for my midterm on my week points for calculus.
amistre64
  • amistre64
there are relatively few cardinal sins in math; the biggest is dividing by a zero
anonymous
  • 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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amistre64
  • amistre64
what values of x make the bottom go zero?
anonymous
  • anonymous
Finding the zero's?
amistre64
  • amistre64
the zeros of the bottom, not the zeros of the function perse
anonymous
  • anonymous
I remember trying to solve for zero
amistre64
  • amistre64
then solve for zeros of the bottom: (3x-1)(2x+5) = 0, for what values of x?
anonymous
  • anonymous
making the equation equal to zero?
anonymous
  • anonymous
yes exactly
amistre64
  • amistre64
recall your multiplication tables .... 0 was one of the simplest ones to remember
amistre64
  • amistre64
given a product of a and b ab = 0 when a=0, or b=0, or both equal 0
anonymous
  • anonymous
Anything times zero is zero lol
amistre64
  • amistre64
yes :) so (3x-1)(2x+5) = 0, when 3x-1 = 0 , or when 2x+5 = 0
anonymous
  • anonymous
4x/((3x-1)(2x+5)) 4x = 0 3x -1 = 0 2x + 5 = 0
amistre64
  • 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
  • amistre64
a function is discontinuous where it is undefined, or where it breaks apart
amistre64
  • amistre64
a fraction is undefined when it has a zero underneath: n/0 is bad
anonymous
  • anonymous
I get that
amistre64
  • 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
  • anonymous
So the numerator is allow to have a zero but the denominator is not allowed
amistre64
  • amistre64
correct
anonymous
  • anonymous
So fill in values for x?
amistre64
  • amistre64
of course, or algebra out some values
amistre64
  • amistre64
im not real sure what process youre thinking of there
amistre64
  • 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
  • anonymous
3x -1 = 0 3x = 0 + 1 or 2x + 5 = 0 2x = 0 + 5
amistre64
  • amistre64
2x = -5 , but yeah
anonymous
  • anonymous
yeah i forgot to flip the sign
amistre64
  • 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
  • amistre64
there are 5 simple properties to algebra, and they can prevent a world of mistakes
amistre64
  • 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
  • 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
  • anonymous
okay so then we try to get x alone
anonymous
  • anonymous
I understand that
amistre64
  • amistre64
yes, by using the properties of algebra, we can isolate the variable
anonymous
  • anonymous
thanks
anonymous
  • anonymous
signs reverse on the other side of the equation
amistre64
  • 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
  • anonymous
okay so the inverse becomes the identity
amistre64
  • amistre64
correct, applying the inverse of a given element, produces the identity element
anonymous
  • anonymous
inverse meaning the opposite and identity the after math
amistre64
  • amistre64
yep
anonymous
  • anonymous
so this function is discontinuous at x = -5/2
amistre64
  • amistre64
that is one of the places yes
amistre64
  • amistre64
the other one is at x=1/3
amistre64
  • 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
  • anonymous
Okay thanks you up for another one with lim?
amistre64
  • 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
  • anonymous
lmao yeah thats how I feel right about now

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