Functions

To solve for the range of the function , we must first understand the behavior of each term: .

Step 1: Simplification and Analysis
The expression simplifies to consider positive exponential terms and and negative exponential terms and .
Observe the nature of the function: As approaches positive or negative infinity, we examine limits individually.

Step 2: Limit Analysis

• As , , , while , . Thus, .
• As , , , while , . The function again grows large due to the dominance of and .

Step 3: Minimum Value Calculation
Since the components and are symmetric and reach minimal values of 2 at due to AM-GM inequality (arithmetic mean-geometric mean inequality), the entire function becomes minimal at this point.

The minimum value of

Step 4: Range Conclusion
Considering the asymptotic behavior and the calculated minimum value, ranges from at least 6 upwards. Therefore, the range is , making option [6,∞) the correct answer.

To find the distinct linear functions that map the interval onto , we need to consider the general form of a linear function which is . For the function to map to , the endpoints of the domain must map to the endpoints of the range:

1. At , should equal 0: .
2. At , should equal 2: .

We have a system of equations:

(1)
(2)

Substituting into :

• Therefore, from .

This gives the linear function . To find the second function, consider reflection over the x-axis to maintain linearity and distinct mapping. Hence, we try . Both functions and satisfy the condition of distinct mappings and linear properties.

Therefore, the functions are: .

We need to determine the domain of the function defined by the equation , where .

1. Interpret the Equation:
The given equation is . We need to find the values of for which is defined, i.e., the domain of . This means we must solve for in terms of and determine the values of that produce a real .

2. Solve for :
Start with the equation:


Isolate :


For to be defined and real, the right-hand side must be positive (since for all real , as ):



Since , divide both sides by :


Thus:


Since , implies the exponent must be negative:

3. Express :
Now, take the natural logarithm on both sides of :



For to be real, the argument of the logarithm must be positive, which we already established as , or . Additionally, since , so the denominator is positive and well-defined.

4. Check Boundary Conditions:
- As , , so , making , and thus , which is still real.
- For , , so is real, and is defined.
- If , then , making undefined (since the logarithm of a non-positive number is not real).

5. Determine the Domain:
The condition ensures , allowing to be defined. There are no other restrictions, as ensures all terms are well-behaved for . Thus, the domain of is all real numbers less than 1.

Final Answer:
The domain of the function defined by the equation (where ) is: