Question

Study the following functions and determine if they are continuous. if not, state where their discontinuities exist and what type

Answer 1

\[f(x)= \frac{ 5 }{ x^4-16}\]

Answer 2

Hint: You can't divide by zero. Is it possible for \(x\) to be such that computing the function would require dividing by zero? If so, what value of \(x\) would cause that?

Answer 3

2

Answer 4

actually, 3

Answer 5

I got a big homework. I just need to understand how you go on about in understanding how it is discontinuous or not

Answer 6

i get the whole indivisible by zero part

Answer 7

there would be a hole if the denominator was zero

Answer 8

Frank, help me out over here haha

Answer 9

You were right the first time: when \(x = 2\) we have \(x^4 = 2^4 = 16\) and the denominator of the formula would be zero. And if the denominator is zero then the formula can't produce an answer, and the function has no value.

Remember the definition of a continuous function: \(f(x)\) is continuous at \(x_0\) if \(lim_{x\rightarrow x_0} = f(x_0)\). Which means, among other things, that the function \(f\) has to have a value at \(x_0\).

So is this particular function continuous at \(x = 2\), or not?

Answer 10

I said 3 so it wouldn't equal to zero

Answer 11

i'm guessing you can't just randomly plug in a number?

Answer 12

At \(x = 3\) the function would have a value. However, that's not what the problem is asking. The problem is asking you if there are places where the function is not continuous, which means places where the function would *not* have a value. So, again, is there a place where this function doesn't have a value, and thus a place where it's not continuous?

Answer 13

yes, at x=2

Answer 14

therefore, the function is discontinuous.

Answer 15

Correct. The function is not continuous at \(x = 2\).

Answer 16

what type of discontinuity would it be?

Answer 17

Remember the three types of discontinuity: 1. Removable discontinuity: The function doesn't happen to have a value at 2 (in this case), but you could figure out what it should be based on the value of the function where \(x\) is close to 2 on either side.
2. Jump discontinuity: The function jumps in value when you go from one side of 2 to the other side. For example: a function where the value of the function is 0 for all \(x \le 2\) and 1 for all \(x > 2\). 3. Infinite discontinuity: The value goes to positive or negative infinity (keeps getting ever larger or smaller) the closer you get to 2.

So which of the three sounds like it would be the case here? hint: If you have a calculator, try computing the function for 2.0001:\[f(2.0001) = \frac{5}{2.0001^4 -16}\]

Answer 18

1562.3828

Answer 19

OK, now try it for 2.00001.

Answer 20

therefore, it would be an infinite discontinuity.

Answer 21

Correct, infinite discontinuity.

Answer 22

because if you're going to keep increasing x, then it's positive infinity. give me a problem with one of the two please :D

Answer 23

i'm learning a lot here.

Answer 24

Oh, one more thing I forgot to add: Is \(x = 2\) the only place where this function is not continuous? Hint: Is there any other value of \(x\) for which \(x^4=16\)?

Answer 25

1

Answer 26

No, \(1^4 = 1 \cdot 1 \cdot 1 \cdot 1 = 1\). Try again :-) Hint: Try negative values of \(x\).

Answer 27

-2

Answer 28

Correct. This function is not continuous at 2 and -2. What type of discontinuity is at -2?

Answer 29

Removal discontinuity.

Answer 30

Removal discontinuity is if the function isn't defined to have a value at -2, but you can figure out what the value of the function should be based on the values of the function when \(x\) is close to -2 on either side.

For example, suppose we define the function \(g(x)\) to have value of \(x + 2\) for \(x < -2\) and a value of \(2x + 4\) for \(x > -2\), but we don't define a value at -2 itself. What number would make sense for a value of the function at \(x=-2\)?

Answer 31

-1

Answer 32

and -3

Answer 33

No, not sure where you got those numbers. But think of it this way: If \(g(x)=x+2\) for \(x<−2\) and \(g(x)=2x+4\) for \(x>−2\), what would the value of \(g\) be at \(x=-2.001\)? What about at \(x = 1.999\)?

Answer 34

Sorry, I meant \(x = -1.999\).

Answer 35

-oo

Answer 36

Sorry, are you talking about the original function \(f(x) = \frac{5}{x^4-16}\)? You're right, that function goes infinite when \(x\) is close to -2. So that would be a "limit of infinity" discontinuity, same as it would be for values of \(x\) close to 2..

Answer 37

i don't really understand the equation you just gave me. How would i go on about to solve it

Answer 38

However, for the second function I gave you, where \(g(x) = x+2\) for \(x < -2\) and \(g(x) = 2x+4\) for \(x > -2\), when \(x\) is close to -2 the function \(g(x)\) is close to 0. So the value of \(g(x)\) at \(x = -2\) should be exactly 0. If the value of \(g(-2)\) is not 0 then the function has a removable discontinuity.

Answer 39

And the way I figured that out is simply by plugging -2 into the function formulas: if \(x = -2\) then \[x+2 = -2+2 = 0\]and\[2x+4 = 2(-2) + 4 = -4+4 = 0\]

Answer 40

ahhh all right, that makes more sense!!

Answer 41

wow, missed the value

Answer 42

Incidentally, the function \(g(x)\) looks something like the following:

Answer 43

this is exceptional teaching. All right, it all adds up to me now.

Answer 44

Glad to hear it. Now it's bedtime for me. Good night!

Answer 45

Thank you so much, I appreciate the patience and effort you put in here. people like you change my life! all right, have a great night! thanks again and hopefully be seing you in the future

Answer 46

Will do...