Question

If the maximum value of the function $f(x) = \alpha - 4x - x^2$ is 1, then the value of $\alpha$ is equal to

• A. -3
• B. 3
• C. -5
• D. 5

Hint:

Alternatively, use differentiation: $f'(x) = -4 - 2x$. Setting $f'(x) = 0$ gives $x = -2$. Substitute this back into $f(x)$ and set equal to 1 to solve for $\alpha$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The function \(f(x)\) is a downward-opening parabola (since the coefficient of \(x^2\) is negative). Its maximum value occurs at its vertex.
Step 2: Key Formula or Approach:
Method 1 (Completing the square): Rewrite as \(f(x) = - (x - h)^2 + k\), where \(k\) is the maximum value.
Method 2 (Calculus): Find where \(f'(x) = 0\), and substitute that \(x\) back into \(f(x)\).
Step 3: Detailed Explanation:
Let's use completing the square.
\[ f(x) = -x^2 - 4x + \alpha \] Factor out the -1 from the \(x\) terms:
\[ f(x) = -(x^2 + 4x) + \alpha \] Complete the square inside the parenthesis by adding and subtracting \((\frac{4}{2})^2 = 4\):
\[ f(x) = -(x^2 + 4x + 4 - 4) + \alpha \] \[ f(x) = -(x + 2)^2 + 4 + \alpha \] The maximum value of this function is achieved when the squared term is zero (at \(x = -2\)).
The maximum value is \(4 + \alpha\).
We are given that the maximum value is 1:
\[ 4 + \alpha = 1 \] \[ \alpha = 1 - 4 = -3 \] Step 4: Final Answer:
The value of \(\alpha\) is -3.