The range of a function is the set of all possible values of the function. If the function f is...

1. Translate the problem requirements: We need to find all possible output values (y-values) that the function \(\mathrm{f(x) = \frac{1}{x^2 + 1}}\) can produce when x takes any real number value. This means finding the range of the function.
2. Analyze the denominator behavior: Examine how \(\mathrm{x^2 + 1}\) behaves for different values of x to understand the constraints on the function's output.
3. Determine the function's maximum value: Find when the denominator is smallest to identify the largest possible function value.
4. Establish the function's lower bound: Analyze what happens as the denominator grows very large to find the limiting behavior of the function.
5. Verify the range boundaries: Confirm that the function can actually achieve values arbitrarily close to both the maximum and minimum bounds.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're looking for. When we talk about the "range" of a function, we're asking: "What are all the possible output values this function can give us?"

Think of it like this: imagine you have a machine (our function) that takes any real number as input and spits out a result using the formula \(\mathrm{f(x) = \frac{1}{x^2 + 1}}\). We want to know what collection of numbers this machine could possibly produce as outputs.

So our job is to find all possible y-values where \(\mathrm{y = \frac{1}{x^2 + 1}}\) for any real number x.

Process Skill: TRANSLATE - Converting the range question into finding all possible output values

2. Analyze the denominator behavior

Let's think about what happens in the denominator: \(\mathrm{x^2 + 1}\).

First, notice something important about \(\mathrm{x^2}\): no matter what real number we put in for x, when we square it, we always get a non-negative result. For example:

• If \(\mathrm{x = 3}\), then \(\mathrm{x^2 = 9}\)
• If \(\mathrm{x = -3}\), then \(\mathrm{x^2 = 9}\) (still positive!)
• If \(\mathrm{x = 0}\), then \(\mathrm{x^2 = 0}\)
• If \(\mathrm{x = 100}\), then \(\mathrm{x^2 = 10,000}\)

So \(\mathrm{x^2 \geq 0}\) for any real number x.

This means \(\mathrm{x^2 + 1 \geq 1}\) for any real number x.

The smallest value the denominator can ever be is 1 (when \(\mathrm{x = 0}\)), and it can grow as large as we want by choosing larger values of \(\mathrm{|x|}\).

3. Determine the function's maximum value

Since our function is \(\mathrm{f(x) = \frac{1}{x^2 + 1}}\), and we're dividing 1 by the denominator, the function will be largest when the denominator is smallest.

From step 2, we know the denominator is smallest when \(\mathrm{x^2 + 1 = 1}\), which happens when \(\mathrm{x = 0}\).

When \(\mathrm{x = 0}\):
\(\mathrm{f(0) = \frac{1}{0^2 + 1} = \frac{1}{1} = 1}\)

So the maximum value our function can achieve is 1.

Let's verify this makes sense: as the denominator gets bigger than 1, we're dividing 1 by bigger and bigger numbers, so our result gets smaller. Therefore, 1 is indeed our maximum.

4. Establish the function's lower bound

Now let's think about what happens when we make x very large (either positive or negative).

If \(\mathrm{x = 10}\): \(\mathrm{f(10) = \frac{1}{100 + 1} = \frac{1}{101} \approx 0.0099}\)
If \(\mathrm{x = 100}\): \(\mathrm{f(100) = \frac{1}{10,000 + 1} = \frac{1}{10,001} \approx 0.0001}\)
If \(\mathrm{x = 1,000}\): \(\mathrm{f(1,000) = \frac{1}{1,000,000 + 1} \approx 0.000001}\)

As x gets larger and larger, \(\mathrm{x^2 + 1}\) gets larger and larger, so \(\mathrm{\frac{1}{x^2 + 1}}\) gets closer and closer to 0.

But notice: the function never actually equals 0, because the denominator \(\mathrm{x^2 + 1}\) is never infinite - it's always a finite positive number.

Also notice: since \(\mathrm{x^2 + 1}\) is always positive, and we're dividing the positive number 1 by it, our function is always positive.

5. Verify the range boundaries

Let's confirm our findings:

• Maximum value: \(\mathrm{f(x) = 1}\) when \(\mathrm{x = 0}\) ✓
• The function approaches but never reaches 0 as \(\mathrm{|x|}\) grows large ✓
• The function is always positive since we're dividing 1 by a positive number ✓
• The function can get arbitrarily close to any value between 0 and 1

For example, if we want \(\mathrm{f(x) = 0.5}\):
\(\mathrm{\frac{1}{x^2 + 1} = 0.5}\)
\(\mathrm{x^2 + 1 = 2}\)
\(\mathrm{x^2 = 1}\)
\(\mathrm{x = \pm 1}\)

Indeed: \(\mathrm{f(1) = f(-1) = \frac{1}{1 + 1} = \frac{1}{2} = 0.5}\) ✓

Process Skill: CONSIDER ALL CASES - Checking that we can achieve values throughout our proposed range

6. Final Answer

The range of \(\mathrm{f(x) = \frac{1}{x^2 + 1}}\) is all positive real numbers that are less than or equal to 1.

In interval notation: \(\mathrm{(0, 1]}\)

Looking at our answer choices, this matches choice E: "all positive real numbers less than or equal to 1."

The answer is E.

Common Faltering Points

Errors while devising the approach

1. Confusing range with domain

Students often mix up "range" (output values) with "domain" (input values). Since the question asks for the range of \(\mathrm{f(x) = \frac{1}{x^2 + 1}}\), some students might think about what x-values are allowed (which is all real numbers) instead of what y-values the function can produce. This confusion leads them to incorrectly select answer choice A (all real numbers).

2. Misunderstanding what "set of all possible values" means

Students may not realize they need to find both the maximum and minimum bounds of the function's output. They might think finding one extreme value is sufficient, missing the critical step of analyzing the function's behavior as x approaches different limits.

Errors while executing the approach

1. Incorrectly analyzing the denominator's minimum value

Students might think that since x can be any real number, \(\mathrm{x^2 + 1}\) can equal zero or be negative. They fail to recognize that \(\mathrm{x^2 \geq 0}\) for all real numbers, which means \(\mathrm{x^2 + 1 \geq 1}\) always. This error leads to incorrect conclusions about the function's maximum value.

2. Assuming the function can equal zero

When analyzing what happens as x gets very large, students might incorrectly conclude that \(\mathrm{f(x)}\) actually reaches zero rather than just approaching it. They don't distinguish between "approaches 0" and "equals 0," leading them to incorrectly include 0 in the range.

3. Missing that the function is always positive

Students might not recognize that since both the numerator (1) and denominator \(\mathrm{(x^2 + 1)}\) are always positive, the function must always be positive. This oversight could lead them to consider negative values as possible outputs.

Errors while selecting the answer

1. Choosing answer D due to boundary confusion

Students who correctly identify that the function ranges from approaching 0 to reaching 1 might select answer choice D ("all real numbers greater than or equal to 0 and less than or equal to 1") because they incorrectly think the function can equal 0, when it can only approach 0 but never reach it.

2. Not noticing the subtle difference between choices D and E

Students might not carefully read that choice D includes 0 ("greater than or equal to 0") while choice E excludes 0 ("positive real numbers"). Since the function never actually equals 0, only choice E is correct, but students rushing through might miss this crucial distinction.