1. **State the problem:** Find the maximum value of the function. Since the function is not explicitly given, let's assume a general function $y=f(x)$ and explain the process to find its maximum. 2. **Formula and rules:** To find the maximum of a function, we use the first derivative test. The critical points occur where the derivative $f'(x)=0$ or is undefined. 3. **Steps:** - Compute the derivative $f'(x)$. - Solve $f'(x)=0$ to find critical points. - Use the second derivative test: if $f''(x)<0$ at a critical point, the function has a local maximum there. 4. **Example:** Suppose $f(x)= -x^2 + 4x + 1$. - Derivative: $f'(x) = -2x + 4$. - Set derivative to zero: $-2x + 4 = 0$. - Solve: $$-2x + 4 = 0 \implies \cancel{-2}x + \cancel{4} = 0 \implies x = 2$$. - Second derivative: $f''(x) = -2$ which is less than zero, so $x=2$ is a maximum. - Maximum value: $f(2) = -(2)^2 + 4(2) + 1 = -4 + 8 + 1 = 5$. 5. **Answer:** The maximum value of the function is $5$ at $x=2$.