https://file.aiquickdraw.com/mathbot/file/1726323227130egr6hbql.jpg

Let's go through each part of the problem step by step. The given function is:

\[
f(x) = e^{x-1} + 2
\]

### Part i: Να αποδείξετε ότι η \( f \) είναι γνησίως αύξουσα.
(Translation: Prove that \( f \) is strictly increasing.)

To show that a function is strictly increasing, we need to show that its derivative is positive for all \( x \).

1. First, compute the derivative of \( f(x) \):
   \[
   f'(x) = \frac{d}{dx}\left(e^{x-1} + 2\right)
   \]
   The derivative of \( e^{x-1} \) is \( e^{x-1} \) (since the derivative of an exponential function remains the same).
   
   Thus, \( f'(x) = e^{x-1} \).
   
2. Since \( e^{x-1} \) is always positive for all real \( x \) (because the exponential function is always positive), we conclude that \( f'(x) > 0 \) for all \( x \in \mathbb{R} \).
   
Therefore, \( f(x) \) is strictly increasing.

### Part ii: Να βρείτε την αντίστροφη \( f^{-1}(x) \).
(Translation: Find the inverse \( f^{-1}(x) \).)

To find the inverse function, we follow these steps:

1. Start with the equation:
   \[
   y = f(x) = e^{x-1} + 2
   \]
   
2. Solve for \( x \) in terms of \( y \):
   \[
   y - 2 = e^{x-1}
   \]
   Take the natural logarithm (ln) of both sides:
   \[
   \ln(y - 2) = x - 1
   \]
   
3. Finally, solve for \( x \):
   \[
   x = \ln(y - 2) + 1
   \]
   
Thus, the inverse function is:
\[
f^{-1}(x) = \ln(x - 2) + 1
\]
Note: The domain of the inverse function is \( x > 2 \) because \( \ln(x - 2) \) requires \( x - 2 > 0 \).

### Part iii: Να λύσετε την εξίσωση \( f(2 - e^{x-1}) = 3 \).
(Translation: Solve the equation \( f(2 - e^{x-1}) = 3 \).)

1. Use the given function \( f(x) = e^{x-1} + 2 \):
   \[
   f(2 - e^{x-1}) = e^{(2 - e^{x-1}) - 1} + 2 = 3
   \]
   
   Simplify:
   \[
   e^{1 - e^{x-1}} + 2 = 3
   \]
   Subtract 2 from both sides:
   \[
   e^{1 - e^{x-1}} = 1
   \]
   
2. The exponential equation \( e^a = 1 \) holds if and only if \( a = 0 \). Therefore:
   \[
   1 - e^{x-1} = 0
   \]
   Solve for \( e^{x-1} \):
   \[
   e^{x-1} = 1
   \]
   
   Taking the natural logarithm of both sides:
   \[
   x - 1 = \ln(1) = 0
   \]
   Therefore:
   \[
   x = 1
   \]

### Part iv: Να λύσετε την ανίσωση \( f(3 - e^{x-1}) < e + 2 \).
(Translation: Solve the inequality \( f(3 - e^{x-1}) < e + 2 \).)

1. Start with the inequality:
   \[
   f(3 - e^{x-1}) = e^{(3 - e^{x-1}) - 1} + 2 < e + 2
   \]
   
   Simplify the inequality:
   \[
   e^{2 - e^{x-1}} + 2 < e + 2
   \]
   Subtract 2 from both sides:
   \[
   e^{2 - e^{x-1}} < e
   \]
   
2. Take the natural logarithm of both sides:
   \[
   2 - e^{x-1} < 1
   \]
   
3. Solve for \( e^{x-1} \):
   \[
   e^{x-1} > 1
   \]
   Take the natural logarithm again:
   \[
   x - 1 > 0
   \]
   Therefore:
   \[
   x > 1
   \]

Thus, the solution to the inequality is \( x > 1 \).

---

Would you like further details or have any questions? 

Here are 5 related questions for further practice:

1. Prove that \( f(x) = \ln(x - 1) \) is strictly increasing.
2. Find the inverse of \( g(x) = 2e^{x+3} - 1 \).
3. Solve the equation \( f(e^{x-2} - 1) = 4 \) where \( f(x) = e^{x-1} + 2 \).
4. Solve the inequality \( e^{x-1} + 2 > 4 \).
5. Investigate the behavior of \( f(x) = e^{x-1} + 2 \) as \( x \to \infty \).

**Tip:** Always check the domain when dealing with logarithmic or exponential functions, as they impose constraints on the values of \( x \).

Given the function f(x) = e^(x-1) + 2, solve the following: 
(i) Prove that f is strictly increasing. 
(ii) Find the inverse f^(-1)(x). 
(iii) Solve the equation: f(2 - e^(x-1)) = 3. 
(iv) Solve the inequality: f(3 - e^(x-1)) < e + 2.

Step-by-step solution for solving the exponential function f(x) = e^(x-1) + 2. Learn how to prove that f is strictly increasing, find its inverse, solve an equation and an inequality involving the function. Suitable for university-level calculus students.

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