To determine where the function \( f(x) = x^2 - 4x + 4 \) is increasing or decreasing, we need to find its first derivative and analyze the critical points. 1. **Find the first derivative of \( f(x) \):** \[ f(x) = x^2 - 4x + 4 \] \[ f'(x) = \frac{d}{dx}(x^2 - 4x + 4) = 2x - 4 \] 2. **Find the critical points:** Critical points occur where the first derivative is zero or undefined. Set \( f'(x) = 0 \): \[ 2x - 4 = 0 \] \[ 2x = 4 \] \[ x = 2 \] 3. **Determine the intervals to test:** The critical point divides the real number line into two intervals: \( (-\infty, 2) \) and \( (2, \infty) \). 4. **Test the sign of \( f'(x) \) in each interval:** - For \( x < 2 \) (e.g., \( x = 0 \)): \[ f'(0) = 2(0) - 4 = -4 \quad (\text{negative}) \] - For \( x > 2 \) (e.g., \( x = 3 \)): \[ f'(3) = 2(3) - 4 = 2 \quad (\text{positive}) \] 5. **Determine where the function is increasing or decreasing:** - \( f'(x) < 0 \) for \( x < 2 \), so \( f(x) \) is decreasing on \( (-\infty, 2) \). - \( f'(x) > 0 \) for \( x > 2 \), so \( f(x) \) is increasing on \( (2, \infty) \). Therefore, the correct answer is: \[ \boxed{E} \]

To determine where the function $f(x) = x^2 - 4x + 4$ is increasing or decreasing, we need to find its first derivative and analyze the critical points.

1. Find the first derivative of $f(x)$:

$f(x) = x^2 - 4x + 4$

The derivative of $x^2$ is $2x$, the derivative of $-4x$ is $-4$, and the derivative of a constant like $4$ is $0$. So,

$f'(x) = \frac{d}{dx}(x^2 - 4x + 4) = 2x - 4$

2. Find the critical points:

Critical points occur where the first derivative is zero or undefined. Set $f'(x) = 0$:

$2x - 4 = 0$

Solving for $x$ gives:

$2x = 4$

$x = 2$

3. Determine the intervals to test:

The critical point divides the real number line into two intervals: $(-\infty, 2)$ and $(2, \infty)$.

4. Test the sign of $f'(x)$ in each interval:

• For $x < 2$ (e.g., $x = 0$):

$f'(0) = 2(0) - 4 = -4 \quad (\text{negative})$

• For $x > 2$ (e.g., $x = 3$):

$f'(3) = 2(3) - 4 = 2 \quad (\text{positive})$

5. Determine where the function is increasing or decreasing:

• $f'(x) < 0$ for $x < 2$, so $f(x)$ is decreasing on $(-\infty, 2)$.

• $f'(x) > 0$ for $x > 2$, so $f(x)$ is increasing on $(2, \infty)$.

Therefore, the function $f(x) = x^2 - 4x + 4$ is decreasing on the interval $(-\infty, 2)$ and increasing on the interval $(2, \infty)$.

To determine where the function $f(x) = x^2 - 4x + 4$ is increasing or decreasing, we need to find its first derivative and analyze the critical points.

1. Find the first derivative of $f(x)$:

The function is $f(x) = x^2 - 4x + 4$. The derivative of $x^2$ is $2x$, the derivative of $-4x$ is $-4$, and the derivative of a constant like $4$ is $0$. So, the first derivative of the function is $f'(x) = 2x - 4$.

2. Find the critical points:

Critical points occur where the first derivative is zero or undefined. To find these points, we set $f'(x) = 0$ and solve for $x$:

$2x - 4 = 0$ $2x = 4$ $x = 2$

So, the critical point is $x = 2$.

3. Determine the intervals to test:

The critical point divides the real number line into two intervals: $(-\infty, 2)$ and $(2, \infty)$.

4. Test the sign of $f'(x)$ in each interval:

• For $x < 2$ (e.g., $x = 0$): $f'(0) = 2(0) - 4 = -4$, which is negative.
• For $x > 2$ (e.g., $x = 3$): $f'(3) = 2(3) - 4 = 2$, which is positive.

5. Determine where the function is increasing or decreasing:

• $f'(x) < 0$ for $x < 2$, so $f(x)$ is decreasing on $(-\infty, 2)$.
• $f'(x) > 0$ for $x > 2$, so $f(x)$ is increasing on $(2, \infty)$.

Therefore, the function $f(x) = x^2 - 4x + 4$ is decreasing on the interval $(-\infty, 2)$ and increasing on the interval $(2, \infty)$.