Why Doesn't the Intermediate Value Theorem Apply to ln(x^2 + 2) on [−2, 2]?

In summary: An important point is that this does NOT tell us that there is no zero between -2 and 2. The intermediate value theorem says that IF f(a) is negative and f(b) is positive then there is a zero between a and b. It says nothing at all about what happens if f(a) and f(b) are both positive or both negative. There might be a zero between a and b or there might not.
  • #1
nycmathdad
74
0
Explain why the Intermediate Value Theorem gives no information about the zeros of the function
f(x) = ln(x^2 + 2) on the interval [−2, 2].

Let me see.

Let x = -2.

f(-2) = ln((-2)^2 + 2)

f(-2) = ln(4 + 2)

f(-2) = ln (6). This is a positive value.

When I let x be 2, I get the same answer.

So, f(-2) = f(2).

So, I conclude by saying that the Intermediate Value Theorem does not apply here because both answers are the same value and both are positive.

However, the textbook answer is different.

Textbook Answer:

"This does not contradict the IVT. f (−2) and f (2) are both positive, so there is no guarantee that the function has a zero in the interval (−2, 2)."
 
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  • #2
Beer soaked ramblings follow.
nycmathdad said:
Explain why the Intermediate Value Theorem gives no information about the zeros of the function
f(x) = ln(x^2 + 2) on the interval [−2, 2].

Let me see.

Let x = -2.

f(-2) = ln((-2)^2 + 2)

f(-2) = ln(4 + 2)

f(-2) = ln (6). This is a positive value.

When I let x be 2, I get the same answer.

So, f(-2) = f(2).

So, I conclude by saying that the Intermediate Value Theorem does not apply here because both answers are the same value and both are positive.

However, the textbook answer is different.

Textbook Answer:

"This does not contradict the IVT. f (−2) and f (2) are both positive, so there is no guarantee that the function has a zero in the interval (−2, 2)."
I see no problem whatsoever.
20210322_145040.jpg
 
  • #3
nycmathdad said:
Explain why the Intermediate Value Theorem gives no information about the zeros of the function
f(x) = ln(x^2 + 2) on the interval [−2, 2].

Let me see.

Let x = -2.

f(-2) = ln((-2)^2 + 2)

f(-2) = ln(4 + 2)

f(-2) = ln (6). This is a positive value.

When I let x be 2, I get the same answer.

So, f(-2) = f(2).

So, I conclude by saying that the Intermediate Value Theorem does not apply here because both answers are the same value and both are positive.

However, the textbook answer is different.

Textbook Answer:

"This does not contradict the IVT. f (−2) and f (2) are both positive, so there is no guarantee that the function has a zero in the interval (−2, 2)."
On the contrary the textbook answer is giving the part of YOUR answer that is important! you said "both are positive". The other part of what you said "both answers are the same value" is not important. What if both answers had been 0?
 
  • #4
Country Boy said:
On the contrary the textbook answer is giving the part of YOUR answer that is important! you said "both are positive". The other part of what you said "both answers are the same value" is not important. What if both answers had been 0?

I'm curious, what if both answers are 0? I say the Intermediate Value Theorem does not apply.
 
  • #5
I don't know what you mean by "both answers are 0". The answer to the question is NOT a number, it is an explanation, a sentence. And both you and the textbook say that the explanation is that "both f(2) and f(-2) are positive"- 0 is not between them.

An important point is that this does NOT tell us that there is no zero between -2 and 2. The intermediate value theorem says that IF f(a) is negative and f(b) is positive then there is a zero between a and b. It says nothing at all about what happens if f(a) and f(b) are both positive or both negative. There might be a zero between a and b or there might not.
 
  • #6
Country Boy said:
I don't know what you mean by "both answers are 0". The answer to the question is NOT a number, it is an explanation, a sentence. And both you and the textbook say that the explanation is that "both f(2) and f(-2) are positive"- 0 is not between them.

An important point is that this does NOT tell us that there is no zero between -2 and 2. The intermediate value theorem says that IF f(a) is negative and f(b) is positive then there is a zero between a and b. It says nothing at all about what happens if f(a) and f(b) are both positive or both negative. There might be a zero between a and b or there might not.

Here is what you said:
"On the contrary the textbook answer is giving the part of YOUR answer that is important! you said "both are positive". The other part of what you said "both answers are the same value" is not important. What if both answers had been 0?"
Read your question at the end.
 

FAQ: Why Doesn't the Intermediate Value Theorem Apply to ln(x^2 + 2) on [−2, 2]?

What is the Intermediate Value Theorem?

The Intermediate Value Theorem is a mathematical theorem that states that if a continuous function has two values, one above and one below a certain value, then there must be at least one point between those two values where the function equals the given value.

Why is the Intermediate Value Theorem important?

The Intermediate Value Theorem is important because it allows us to prove the existence of solutions to equations that we cannot solve algebraically. It also helps us understand the behavior of continuous functions and their graphs.

How is the Intermediate Value Theorem used in real life?

The Intermediate Value Theorem is used in many real-life applications, such as in economics to prove the existence of equilibrium points, in physics to prove the existence of solutions to certain equations, and in computer science to prove the existence of roots for algorithms.

What are the limitations of the Intermediate Value Theorem?

The Intermediate Value Theorem only applies to continuous functions, which means that there cannot be any breaks or holes in the graph. It also does not tell us the exact location of the solution, only that it exists between two given points.

Can the Intermediate Value Theorem be generalized to higher dimensions?

Yes, the Intermediate Value Theorem can be extended to higher dimensions, known as the Intermediate Value Theorem for Multivariable Functions. This theorem states that if a function is continuous on a closed and bounded region in n-dimensional space, then it must take on every value between its minimum and maximum values at least once.

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