Question

Maximum of \( f(x) = \alpha - 4x - x^2 \), find \(\alpha = ?\)

Hint:

Remember: The maximum or minimum of a quadratic function occurs at the vertex, which can be found by setting the first derivative equal to zero.

Answer 1

Step 1: Differentiate the given function to find critical points.
The given function is \( f(x) = \alpha - 4x - x^2 \). To find the maximum or minimum, we first differentiate the function with respect to \(x\): \[ f'(x) = -4 - 2x \]
Step 2: Set the derivative equal to zero to find the critical point.
For a maximum or minimum, set \(f'(x) = 0\): \[ -4 - 2x = 0 \quad \Rightarrow \quad x = -2 \]
Step 3: Find the value of \(\alpha\).
Since the maximum value of \( f(x) \) is given as 1, substitute \(x = -2\) into the original function: \[ f(-2) = \alpha - 4(-2) - (-2)^2 = \alpha + 8 - 4 = \alpha + 4 \] We are given that the maximum value is 1, so: \[ \alpha + 4 = 1 \quad \Rightarrow \quad \alpha = -3 \] Thus, the value of \(\alpha\) is: \[ \boxed{-3} \]