Let's go through the problem step by step.

Part (a) - Show that $f$ is a surjection for $a = 0$.

Given:

• Function $f: [a, \infty) \to [1, \infty)$ defined by $f(x) = x^2 - 4x + 5$.
• $a = 0$.

To show $f$ is surjective, we need to prove that for every $y \in [1, \infty)$, there exists an $x \in [0, \infty)$ such that $f(x) = y$.

1. Rewrite $f(x)$:

   We can rewrite $f(x) = x^2 - 4x + 5$ by completing the square: $f(x) = (x^2 - 4x + 4) + 1 = (x - 2)^2 + 1$ So, $f(x) = (x - 2)^2 + 1$.

2. Range of $f(x)$:

   • Since $(x - 2)^2 \geq 0$, the smallest value $f(x)$ can attain is 1 (when $x = 2$).
   • Therefore, $f(x)$ maps to $[1, \infty)$.

3. Find $x$ for a given $y$:

   To solve $y = f(x) = (x - 2)^2 + 1$ for $x$: $y - 1 = (x - 2)^2$ $x - 2 = \pm \sqrt{y - 1}$ $x = 2 \pm \sqrt{y - 1}$

   For $x$ to be in the domain $[0, \infty)$, we require $2 - \sqrt{y - 1} \geq 0$, which holds for all $y \geq 1$ because $y \geq 1$.

   Hence, $f(x)$ is surjective for $a = 0$.

Part (b) - Determine the smallest value of $a$ such that $f$ is injective.

To show $f$ is injective, we need to find the smallest $a$ such that if $f(x_1) = f(x_2)$, then $x_1 = x_2$.

1. Analyze $f(x)$:

   The function $f(x) = (x - 2)^2 + 1$ is a quadratic function, and it's symmetric around $x = 2$.

2. Derivative of $f(x)$:

   $f'(x) = 2(x - 2)$.

   • $f'(x) > 0$ for $x > 2$, so $f(x)$ is increasing for $x > 2$.
   • $f'(x) < 0$ for $x < 2$, so $f(x)$ is decreasing for $x < 2$.

   The function $f(x)$ is injective if we restrict its domain to either $x \geq 2$ or $x \leq 2$.

3. Smallest $a$ for injectivity:

   The smallest value of $a$ such that $f(x)$ is injective is $a = 2$.

Part (c) - Determine $f^{-1}$ for $a = 2$.

Given $a = 2$, the function $f(x)$ is injective on $[2, \infty)$.

To find the inverse function $f^{-1}(y)$ such that $f(f^{-1}(y)) = y$:

1. Start with: $y = (x - 2)^2 + 1$

2. Solve for $x$: $y - 1 = (x - 2)^2$ $x - 2 = \sqrt{y - 1} \quad (\text{since we are on the interval } [2, \infty), x \geq 2)$ $x = 2 + \sqrt{y - 1}$

So the inverse function is: $f^{-1}(y) = 2 + \sqrt{y - 1}, \quad y \geq 1$

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Summary:

(a) The function $f(x)$ is surjective for $a = 0$.

(b) The smallest value of $a$ such that $f(x)$ is injective is $a = 2$.

(c) The inverse function for $a = 2$ is $f^{-1}(y) = 2 + \sqrt{y - 1}$.

Would you like any further details or explanations?

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Related Questions:

1. What is the importance of surjectivity and injectivity in defining inverse functions?
2. How does completing the square help in analyzing quadratic functions?
3. What is the geometric interpretation of the injectivity of a function?
4. Can a function be bijective without being either injective or surjective?
5. How would the analysis change if the domain of $f(x)$ were $(-\infty, b]$ instead of $[a, \infty)$?

Tip: When determining if a function is injective, check where its derivative changes sign; injectivity often occurs where the function is either entirely non-increasing or non-decreasing.