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 August 12th, 2017, 07:15 PM #1 Newbie   Joined: Aug 2017 From: Davao Posts: 9 Thanks: 0 Guyss pleasee help me How to find the points of discontinuity of these functions? 1. x²+3/x²-16 2. 3x+2/x²-6x+9 3.3(x+2)³/x³+4x²+x-6 4.x²-3x/x²+2x²+5x 5.x²+2x-1/4x⁴+4x³+3x²-x-1
 August 12th, 2017, 07:27 PM #2 Senior Member   Joined: May 2016 From: USA Posts: 1,047 Thanks: 430 Hmm I thought I had answered this. Look at item number 1. Are there one or more values of x where the function is not defined? Why is it not defined there? What does that mean? By the way, you may need parentheses in one or more of your questions. For example, is the first one $x^2 - \dfrac{3}{x^2} - 16$? That is what you wrote
 August 12th, 2017, 08:05 PM #3 Global Moderator   Joined: Dec 2006 Posts: 19,162 Thanks: 1638 (2). x² - 6x + 9 = (x - 3)² (3). x³ + 4x² + x - 6 = (x + 3)(x + 2)(x - 1) (4). x² + 2x² + 5x = x(3x + 5) (5). 4x^4 + 4x³ + 3x² - x - 1 = (2x + 1)(2x - 1)(x² + x + 1) In each case, for what values of x is the denominator zero?
 August 12th, 2017, 08:17 PM #4 Newbie   Joined: Aug 2017 From: Davao Posts: 9 Thanks: 0 The given for #1 is x²+3 divided by x²-16
 August 12th, 2017, 09:01 PM #5 Global Moderator   Joined: Dec 2006 Posts: 19,162 Thanks: 1638 Is there a typing error in #4?
 August 12th, 2017, 09:29 PM #6 Newbie   Joined: Aug 2017 From: Davao Posts: 9 Thanks: 0 No, the given for #4 is correct. Btw how to know the points of discontinuity?
August 13th, 2017, 04:02 AM   #7
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Quote:
 Originally Posted by Grntxavier Btw how to know the points of discontinuity?
... set each denominator equal to zero and solve for the values of x that make each rational function undefined.

August 13th, 2017, 04:11 AM   #8
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Quote:
 Originally Posted by Grntxavier How to find the points of discontinuity of these functions? 1. x²+3/x²-16 2. 3x+2/x²-6x+9 3.3(x+2)³/x³+4x²+x-6 4.x²-3x/x²+2x²+5x 5.x²+2x-1/4x⁴+4x³+3x²-x-1
Actually none of these functions have any discontinuities, if you use the definition that 99.999% of the mathematical world uses. I don't doubt your teacher doesn't know this however.

Last edited by Micrm@ss; August 13th, 2017 at 04:17 AM.

 August 13th, 2017, 04:55 AM #9 Newbie   Joined: Aug 2017 From: Davao Posts: 9 Thanks: 0 thank you so much guys big help
August 14th, 2017, 02:52 PM   #10
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Quote:
 Originally Posted by Micrm@ss Actually none of these functions have any discontinuities, if you use the definition that 99.999% of the mathematical world uses. I don't doubt your teacher doesn't know this however.
I imagine that you mean that $\displaystyle f(x)= \frac{x^2+3}{x^2- 16}= \frac{x^2+ 3}{(x- 4)(x+ 4)}$ is not defined at x= 4 and x= -4 so cannot be said to be "discontinuous" there.

However the definition I have always used is
"A function is continuous at x= a if and only if:
a) f(a) exists
b) $\displaystyle \lim_{x\to a} f(x)= f(a)$

I would say that a function is discontinuous at points where the function value does not exist. Perhaps you would say it is neither continuous nor discontinuous there.

Last edited by Country Boy; August 14th, 2017 at 02:55 PM.

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