The problem asks for the maximum value of the function:

$f(x) = -x + 2\sqrt{x} + 3$

To find the maximum value, we'll take the derivative of $f(x)$, set it equal to zero, and solve for $x$. This will help us identify the critical points.

Step 1: Find the derivative of $f(x)$

We differentiate each term:

• The derivative of $-x$ is $-1$.
• The derivative of $2\sqrt{x}$ is $\frac{1}{\sqrt{x}}$ (using the chain rule).
• The derivative of the constant 3 is 0.

Thus, the derivative of $f(x)$ is:

$f'(x) = -1 + \frac{1}{\sqrt{x}}$

Step 2: Set the derivative equal to zero

To find the critical points, set $f'(x) = 0$:

$-1 + \frac{1}{\sqrt{x}} = 0$

Solve for $\sqrt{x}$:

$\frac{1}{\sqrt{x}} = 1$

$\sqrt{x} = 1$

Square both sides:

$x = 1$

Step 3: Check if it is a maximum

We now check the second derivative, or we can analyze the behavior of the function to confirm this is a maximum.

• For $x = 1$, substitute into the original function $f(x)$:

$f(1) = -1 + 2\sqrt{1} + 3 = -1 + 2 + 3 = 4$

Step 4: Conclusion

The maximum value of the function $f(x) = -x + 2\sqrt{x} + 3$ is 4, which corresponds to option $\boxed{1}$.

Would you like a more detailed breakdown of any step, or do you have any further questions?

Related Questions:

1. What is the minimum value of the function $f(x) = -x + 2\sqrt{x} + 3$?
2. How do you find critical points using the first derivative test?
3. What is the importance of the second derivative in determining maxima or minima?
4. How can you use limits to analyze the behavior of functions at infinity?
5. What happens to the function $f(x) = -x + 2\sqrt{x} + 3$ as $x \to \infty$?

Tip:

Always double-check the domain of your function when dealing with square roots. In this case, $x \geq 0$ since $\sqrt{x}$ is defined only for non-negative values of $x$.