Question

Find the interval in which the given function \( f(x) = x^2 - 4x + 6 \) is (i) Increasing (ii) Decreasing.

Hint:

For a quadratic function \( ax^2+bx+c \), the vertex at \( x=-b/2a \) is the turning point. This is the critical point where the function changes from decreasing to increasing (if a>0) or increasing to decreasing (if a<0). You can use this to quickly find the intervals.

Answer 1

Step 1: Understanding the Concept:
A function is increasing where its first derivative is positive (\( f'(x)>0 \)) and decreasing where its first derivative is negative (\( f'(x)<0 \)). The points where the derivative is zero or undefined are the critical points, which are the boundaries of these intervals.
Step 2: Key Formula or Approach:
1. Find the first derivative of the function, \( f'(x) \).
2. Find the critical points by setting \( f'(x) = 0 \).
3. Test the sign of \( f'(x) \) in the intervals created by the critical points.
Step 3: Detailed Explanation or Calculation:
The given function is \( f(x) = x^2 - 4x + 6 \).
1. Find the first derivative:
\[ f'(x) = \frac{d}{dx}(x^2 - 4x + 6) = 2x - 4 \] 2. Find the critical point:
Set the derivative equal to zero: \[ 2x - 4 = 0 \] \[ 2x = 4 \] \[ x = 2 \] The critical point is at \( x=2 \). This point divides the number line into two intervals: \( (-\infty, 2) \) and \( (2, \infty) \).
3. Test the intervals:
- Interval 1: \( (-\infty, 2) \)
Choose a test value in this interval, for example, \( x=0 \). \[ f'(0) = 2(0) - 4 = -4 \] Since \( f'(x) \) is negative in this interval, the function \( f(x) \) is decreasing on \( (-\infty, 2) \).
- Interval 2: \( (2, \infty) \)
Choose a test value in this interval, for example, \( x=3 \). \[ f'(3) = 2(3) - 4 = 6 - 4 = 2 \] Since \( f'(x) \) is positive in this interval, the function \( f(x) \) is increasing on \( (2, \infty) \).
Step 4: Final Answer:
(i) The function is increasing on the interval \( (2, \infty) \).
(ii) The function is decreasing on the interval \( (-\infty, 2) \).