For which of the following values of x the inequality 3 `(x^2 - 4x + 4) lt` x gets satisfied ?

Options

A. ` x lt 3, x lt 4/3`
B. `x gt -3 , x lt -4/3`
C. `4/3 lt x lt 3`
D. `4/3 le x le 3`

Answer

To solve the inequality \( 3(x^2 - 4x + 4) < x \), we will follow these steps:

### Step 1: Rewrite the inequality
Start with the given inequality:
\[ 3(x^2 - 4x + 4) < x \]

### Step 2: Expand the left side
Distribute the 3 on the left side:
\[ 3x^2 - 12x + 12 < x \]

### Step 3: Move all terms to one side
Rearranging the inequality gives:
\[ 3x^2 - 12x + 12 - x < 0 \]
This simplifies to:
\[ 3x^2 - 13x + 12 < 0 \]

### Step 4: Factor the quadratic expression
Next, we need to factor the quadratic \( 3x^2 - 13x + 12 \). We look for two numbers that multiply to \( 3 \times 12 = 36 \) and add to \(-13\). The numbers \(-9\) and \(-4\) work:
\[ 3x^2 - 9x - 4x + 12 < 0 \]
Grouping gives:
\[ 3x(x - 3) - 4(x - 3) < 0 \]
Factoring out \((x - 3)\):
\[ (3x - 4)(x - 3) < 0 \]

### Step 5: Find the critical points
Set each factor to zero to find critical points:
1. \( 3x - 4 = 0 \) gives \( x = \frac{4}{3} \)
2. \( x - 3 = 0 \) gives \( x = 3 \)

### Step 6: Test intervals
We will test intervals determined by the critical points \( \frac{4}{3} \) and \( 3 \):
- Interval 1: \( (-\infty, \frac{4}{3}) \)
- Interval 2: \( (\frac{4}{3}, 3) \)
- Interval 3: \( (3, \infty) \)

Choose test points from each interval:
1. For \( x = 0 \) in Interval 1:
\[ (3(0) - 4)(0 - 3) = (-4)(-3) = 12 > 0 \]
2. For \( x = 2 \) in Interval 2:
\[ (3(2) - 4)(2 - 3) = (6 - 4)(-1) = 2(-1) = -2 < 0 \]
3. For \( x = 4 \) in Interval 3:
\[ (3(4) - 4)(4 - 3) = (12 - 4)(1) = 8 > 0 \]

### Step 7: Determine the solution
The inequality \( (3x - 4)(x - 3) < 0 \) is satisfied in the interval:
\[ \left( \frac{4}{3}, 3 \right) \]

### Conclusion
Thus, the values of \( x \) that satisfy the inequality \( 3(x^2 - 4x + 4) < x \) are in the interval \( \left( \frac{4}{3}, 3 \right) \).