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 $x$.