Find the range of f(x), where `f(x)=cos(sin(In(x^2+e)/(x^2+1)))+sin(cos(In(x^2+e)/(x^2+1))).`

Answer

To find the range of the function \( f(x) = \cos(\sin(\ln(\frac{x^2 + e}{x^2 + 1}))) + \sin(\cos(\ln(\frac{x^2 + e}{x^2 + 1}))) \), we will break down the problem step by step.

### Step 1: Simplify the inner function
First, we need to analyze the expression \( \ln\left(\frac{x^2 + e}{x^2 + 1}\right) \).

**Hint:** To simplify, consider the behavior of the function as \( x \) approaches 0 and infinity.

### Step 2: Find the limits of \( \frac{x^2 + e}{x^2 + 1} \)
- As \( x \to 0 \):
\[ \frac{x^2 + e}{x^2 + 1} = \frac{0 + e}{0 + 1} = e \]
- As \( x \to \infty \):
\[ \frac{x^2 + e}{x^2 + 1} = \frac{x^2(1 + \frac{e}{x^2})}{x^2(1 + \frac{1}{x^2})} \to 1 \]

Thus, the function \( \frac{x^2 + e}{x^2 + 1} \) ranges from 1 to \( e \).

### Step 3: Find the range of \( \ln\left(\frac{x^2 + e}{x^2 + 1}\right) \)
Now, we need to find the range of \( \ln(t) \) where \( t \) ranges from 1 to \( e \):
\[ \ln(1) = 0 \quad \text{and} \quad \ln(e) = 1 \]
Thus, the range of \( \ln\left(\frac{x^2 + e}{x^2 + 1}\right) \) is \( [0, 1] \).

### Step 4: Analyze \( \sin \) and \( \cos \) functions
Next, we need to evaluate:
- \( \sin(\ln(t)) \) where \( t \in [1, e] \)
- \( \cos(\ln(t)) \) where \( t \in [1, e] \)

**Hint:** Remember that the sine function is increasing in the interval \( [0, 1] \) and the cosine function is decreasing.

### Step 5: Find the range of \( \sin(\ln(t)) \)
Since \( \ln(t) \) ranges from 0 to 1:
\[ \sin(0) = 0 \quad \text{and} \quad \sin(1) \approx 0.8415 \]
Thus, the range of \( \sin(\ln(t)) \) is \( [0, \sin(1)] \).

### Step 6: Find the range of \( \cos(\ln(t)) \)
For \( \cos(\ln(t)) \):
\[ \cos(0) = 1 \quad \text{and} \quad \cos(1) \approx 0.5403 \]
Thus, the range of \( \cos(\ln(t)) \) is \( [\cos(1), 1] \).

### Step 7: Combine the results
Now we can evaluate \( f(x) \):
\[ f(x) = \cos(\sin(\ln(t))) + \sin(\cos(\ln(t))) \]
We need to find the combined range of these two functions.

**Hint:** Use the extreme values from the ranges of \( \sin \) and \( \cos \) to find the overall range of \( f(x) \).

### Step 8: Determine the final range
The minimum value occurs when both \( \sin \) and \( \cos \) are at their minimum values:
- Minimum of \( f(x) \) is \( 0 + \sin(\cos(1)) \).
- Maximum of \( f(x) \) is \( \cos(0) + \sin(1) = 1 + \sin(1) \).

Thus, the range of \( f(x) \) is:
\[ \left[\sin(\cos(1)), 1 + \sin(1)\right] \]

### Final Answer:
The range of \( f(x) \) is \( \left[\sin(\cos(1)), 1 + \sin(1)\right] \).