Question

Find the interval where the function \( f(x) = x^2 - 4x + 6 \) is strictly decreasing.

• A. \( (-\infty, 2) \)
• B. \( (2, \infty) \)
• C. \( (-\infty, 4) \)
• D. \( [-2, 2] \)

Hint:

Always double-check whether the question specifies "strictly decreasing" (\( f'(x) < 0 \)) or simply "decreasing" (\( f'(x) \le 0 \)). Strictly decreasing requires using open parentheses, while standard decreasing uses a closed bracket at the critical point.

Answer 1

The Correct Option is A

Concept: A continuous function \( f(x) \) is strictly decreasing on any interval where its first derivative remains strictly negative: \[ f'(x) < 0 \] Conversely, it is strictly increasing across zones where the first derivative remains strictly positive (\( f'(x) > 0 \)).

Step 1: Calculate the first derivative of the function \( f'(x) \).
Given the quadratic function \( f(x) = x^2 - 4x + 6 \), apply the power rule for derivatives: \[ f'(x) = \frac{d}{dx}(x^2) - \frac{d}{dx}(4x) + \frac{d}{dx}(6) = 2x - 4 \]

Step 2: Set up the inequality for the strictly decreasing condition.
Set the derivative expression to be strictly less than zero: \[ 2x - 4 < 0 \] Isolate the variable term by moving constants across the inequality sign: \[ 2x < 4 \quad \Rightarrow \quad x < 2 \]

Step 3: Convert the solution into standard interval notation.
The condition \( x < 2 \) includes all real numbers from negative infinity up to but not including 2. Expressed as an open interval: \[ x \in (-\infty, 2) \]