1. The problem asks for a function that has a maximum at $x=12$ and includes an exponential component. 2. A common way to create a function with a maximum at a specific point is to use a quadratic term that opens downward, such as $-(x - h)^2$, where $h$ is the $x$-coordinate of the maximum. 3. To include an exponential function, we can multiply or add an exponential term like $e^x$ or $e^{-x}$. 4. One simple function that satisfies these conditions is: $$f(x) = - (x - 12)^2 + e^x$$ Here, the quadratic term $-(x - 12)^2$ has its maximum at $x=12$, and the exponential term $e^x$ is included. 5. To verify the maximum at $x=12$, consider the derivative: $$f'(x) = -2(x - 12) + e^x$$ 6. At $x=12$: $$f'(12) = -2(12 - 12) + e^{12} = 0 + e^{12} > 0$$ Since the derivative is positive, $x=12$ is not a maximum for this function as is. 7. To ensure a maximum at $x=12$, we can adjust the function to: $$f(x) = -e^{-(x - 12)^2}$$ This function has a maximum at $x=12$ because the exponent is zero there, and the negative sign ensures a peak. 8. The derivative is: $$f'(x) = -e^{-(x - 12)^2} \cdot (-2(x - 12)) = 2(x - 12)e^{-(x - 12)^2}$$ 9. At $x=12$, $f'(12) = 0$, and the second derivative is negative, confirming a maximum. Final answer: $$f(x) = -e^{-(x - 12)^2}$$