Solve the inequality, `(3x^(2) - 7x +8)/(x^(2) + 1) le 2`

Answer

To solve the inequality \(\frac{3x^2 - 7x + 8}{x^2 + 1} \leq 2\), we will follow these steps:

### Step 1: Rewrite the inequality
We start by rewriting the inequality:
\[ \frac{3x^2 - 7x + 8}{x^2 + 1} \leq 2 \]

### Step 2: Clear the fraction
Multiply both sides by \(x^2 + 1\) (which is always positive since \(x^2 + 1 > 0\) for all \(x\)):
\[ 3x^2 - 7x + 8 \leq 2(x^2 + 1) \]

### Step 3: Expand the right side
Expanding the right side gives:
\[ 3x^2 - 7x + 8 \leq 2x^2 + 2 \]

### Step 4: Move all terms to one side
Subtract \(2x^2 + 2\) from both sides:
\[ 3x^2 - 7x + 8 - 2x^2 - 2 \leq 0 \]
This simplifies to:
\[ x^2 - 7x + 6 \leq 0 \]

### Step 5: Factor the quadratic
Now we factor the quadratic:
\[ (x - 6)(x - 1) \leq 0 \]

### Step 6: Find the critical points
The critical points from the factors are \(x = 6\) and \(x = 1\).

### Step 7: Test intervals
We will test the intervals determined by the critical points:
1. \(x < 1\)
2. \(1 \leq x \leq 6\)
3. \(x > 6\)

- **For \(x < 1\)** (e.g., \(x = 0\)):
\((0 - 6)(0 - 1) = 6 > 0\) (not part of the solution)

- **For \(1 \leq x \leq 6\)** (e.g., \(x = 3\)):
\((3 - 6)(3 - 1) = (-3)(2) = -6 \leq 0\) (part of the solution)

- **For \(x > 6\)** (e.g., \(x = 7\)):
\((7 - 6)(7 - 1) = (1)(6) = 6 > 0\) (not part of the solution)

### Step 8: Write the solution
The solution to the inequality is:
\[ 1 \leq x \leq 6 \]
In interval notation, this is:
\[ x \in [1, 6] \]