For all numbers x, let `f(x) = x^(2) - 6x + 8` and g(x) = x- 2 . For how many integer values of x is `g(x) ge f(x)` ?

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to find the integer values of \( x \) such that \( g(x) \geq f(x) \), where \( f(x) = x^2 - 6x + 8 \) and \( g(x) = x - 2 \).

### Step 1: Set up the inequality
We start with the inequality:
\[ g(x) \geq f(x) \]
Substituting the expressions for \( g(x) \) and \( f(x) \):
\[ x - 2 \geq x^2 - 6x + 8 \]

### Step 2: Rearrange the inequality
Rearranging the inequality gives:
\[ 0 \geq x^2 - 6x + 8 - (x - 2) \]
This simplifies to:
\[ 0 \geq x^2 - 7x + 10 \]
or equivalently:
\[ x^2 - 7x + 10 \leq 0 \]

### Step 3: Factor the quadratic expression
Next, we factor the quadratic:
\[ x^2 - 7x + 10 = (x - 5)(x - 2) \]
Thus, we rewrite the inequality:
\[ (x - 5)(x - 2) \leq 0 \]

### Step 4: Determine the critical points
The critical points from the factors are \( x = 5 \) and \( x = 2 \). These points divide the number line into intervals:
1. \( (-\infty, 2) \)
2. \( (2, 5) \)
3. \( (5, \infty) \)

### Step 5: Test the intervals
We will test each interval to see where the product \( (x - 5)(x - 2) \) is less than or equal to zero.

- For \( x < 2 \) (e.g., \( x = 0 \)):
\[ (0 - 5)(0 - 2) = 10 > 0 \]

- For \( 2 < x < 5 \) (e.g., \( x = 3 \)):
\[ (3 - 5)(3 - 2) = (-2)(1) = -2 < 0 \]

- For \( x > 5 \) (e.g., \( x = 6 \)):
\[ (6 - 5)(6 - 2) = (1)(4) = 4 > 0 \]

### Step 6: Include the endpoints
The inequality \( (x - 5)(x - 2) \leq 0 \) holds true in the interval \( [2, 5] \).

### Step 7: Identify integer solutions
The integer values of \( x \) in the interval \( [2, 5] \) are:
- \( 2 \)
- \( 3 \)
- \( 4 \)
- \( 5 \)

Thus, there are **4 integer values** of \( x \) that satisfy the inequality \( g(x) \geq f(x) \).

### Final Answer
The number of integer values of \( x \) such that \( g(x) \geq f(x) \) is **4**.

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