A line and a circle?

Upon inspection, we see that the function does not yield real values if $x$ is less than -3 or greater than 3. This defines the domain of the given function.

The given function is:

$y = x + 4 +\sqrt{9-x^2}$

Computing its derivative yields:

$\frac{dy}{dx} = 1 - \frac{x}{\sqrt{9-x^2}}$

The derivative is greater than zero as long as x is less than $3/\sqrt{2}$ . This implies that the function increases as x increases from a value of $x=-3$ upto $x=3/\sqrt{2}$ .

This also implies that the maximum value of the function is at $x_p = 3/\sqrt{2}$ .

This maximum value is $y_{max} = 3\sqrt{2} + 4$ . Now, the function decreases beyond and before $x_p$ .

Computing the function at $x=-3$ and $x=3$ gives the result that the value of the function is unity at $x=-3$ .

This is lower than the value the function attains at $x=3$ . This implies that the value of the function is limited between 1 and $y_{max}$ . Hence we have the required answer of ${1 + y_{max} = 3\sqrt{2} + 5}$

Given that $y = x+4+\sqrt{9-x^2}$ , we need to find $a = \min(y)$ and $b = \max(y)$ . We note that the domain where $y$ is real is $x \in [-3,3]$ . Also note that $\min(x)=-3$ and $\min \left(\sqrt{9-x^2}\right) = 0$ occur when $x = -3$ . Therefore $y$ is minimum when $x=-3$ or $a = \min(y) = - 3+4+0 = 1$ .

Assuming the $b = \max(y)$ occurs when $x > 0$ and $\sqrt{9-x^2} > 0$ , then we can apply Cauchy-Schwarz inequality as follows:

$\begin{aligned} \left(x + \sqrt{9-x^2}\right)^2 & \le 2 \left(x^2 + 9 - x^2\right) = 18 & \small \color{#3D99F6} \text{Equality occurs when }x = \sqrt{9-x^2} \\ \implies x + \sqrt{9-x^2} & \le 3 \sqrt 2 & \small \color{#3D99F6} \text{or }x = \frac 3{\sqrt 2}> 0 \text{ as assumed.} \\ \implies y & \le 3\sqrt 2 + 4 \approx 8.2426 \end{aligned}$

Therefore, the range of $y \in [1, 8.2426]$ , $\implies a + b \approx \boxed{9.243}$ .