Question

Find the maximum value of $f(x) = -x^2 + 4x + 1$

• A. \( 3 \)
• B. \( 4 \)
• C. \( 5 \)
• D. \( 6 \)

Hint:

For any quadratic equation form \(f(x) = ax^2 + bx + c\) where \(a < 0\), the vertex occurs at \(x = -\frac{b}{2a}\) and the maximum value is given directly by \( \frac{4ac - b^2}{4a} \). Substituting our coefficients: \( \frac{4(-1)(1) - (4)^2}{4(-1)} = \frac{-4 - 16}{-4} = \frac{-20}{-4} = 5 \).

Answer 1

The Correct Option is C

Concept: To locate the absolute maximum point of a smooth polynomial function, we apply calculus minimization/maximization principles through the First Derivative Test:

• Find the critical points where the first derivative equals zero: \( f'(x) = 0 \)
• Use the Second Derivative Test to confirm a maximum: if \( f''(x) < 0 \), the point is a local maximum.

Alternatively, this can be solved algebraically by completing the square on the quadratic function.

Step 1: Finding the critical value using differentiation.
Given function: \[ f(x) = -x^2 + 4x + 1 \] Differentiate with respect to \(x\): \[ f'(x) = \frac{d}{dx}(-x^2 + 4x + 1) = -2x + 4 \] Set the first derivative to zero to locate the critical turning point: \[ -2x + 4 = 0 \quad \Rightarrow \quad 2x = 4 \quad \Rightarrow \quad x = 2 \] Let's check the second derivative to confirm the behavior: \[ f''(x) = \frac{d}{dx}(-2x + 4) = -2 \] Since \(f''(2) = -2 < 0\), the function reaches a local maximum at \(x = 2\).

Step 2: Calculating the maximum value.
Substitute the critical coordinate position \(x = 2\) back into the original function equation: \[ f(2) = -(2)^2 + 4(2) + 1 \] \[ f(2) = -4 + 8 + 1 = 5 \] Thus, the maximum value of the function is \(5\).