Solve `1le(3x^(2)-7x+8)/(x^(2)+1)le2`.

Answer

To solve the inequality \( 1 \leq \frac{3x^2 - 7x + 8}{x^2 + 1} \leq 2 \), we can break it down into two separate inequalities:

1. \( \frac{3x^2 - 7x + 8}{x^2 + 1} \geq 1 \)
2. \( \frac{3x^2 - 7x + 8}{x^2 + 1} \leq 2 \)

### Step 1: Solve the first inequality

Starting with the first inequality:

\[ \frac{3x^2 - 7x + 8}{x^2 + 1} \geq 1 \]

Multiply both sides by \( x^2 + 1 \) (which is always positive):

\[ 3x^2 - 7x + 8 \geq x^2 + 1 \]

Rearranging gives:

\[ 3x^2 - x^2 - 7x + 8 - 1 \geq 0 \]

This simplifies to:

\[ 2x^2 - 7x + 7 \geq 0 \]

### Step 2: Solve the quadratic inequality

Next, we need to find the roots of the quadratic equation \( 2x^2 - 7x + 7 = 0 \) using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 2, b = -7, c = 7 \):

\[ x = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 2 \cdot 7}}{2 \cdot 2} \]

Calculating the discriminant:

\[ 49 - 56 = -7 \]

Since the discriminant is negative, there are no real roots, and the quadratic \( 2x^2 - 7x + 7 \) is always positive (as the coefficient of \( x^2 \) is positive).

Thus,

\[ 2x^2 - 7x + 7 \geq 0 \text{ for all } x. \]

### Step 3: Solve the second inequality

Now, we solve the second inequality:

\[ \frac{3x^2 - 7x + 8}{x^2 + 1} \leq 2 \]

Again, multiply both sides by \( x^2 + 1 \):

\[ 3x^2 - 7x + 8 \leq 2(x^2 + 1) \]

This simplifies to:

\[ 3x^2 - 7x + 8 \leq 2x^2 + 2 \]

Rearranging gives:

\[ 3x^2 - 2x^2 - 7x + 8 - 2 \leq 0 \]

This simplifies to:

\[ x^2 - 7x + 6 \leq 0 \]

### Step 4: Factor the quadratic

Now we factor the quadratic:

\[ x^2 - 7x + 6 = (x - 6)(x - 1) \leq 0 \]

### Step 5: Determine the intervals

To find the intervals where this inequality holds, we can test the intervals defined by the roots \( x = 1 \) and \( x = 6 \):

- For \( x < 1 \): Choose \( x = 0 \) → \( (0 - 6)(0 - 1) = 6 > 0 \)
- For \( 1 < x < 6 \): Choose \( x = 2 \) → \( (2 - 6)(2 - 1) = -4 < 0 \)
- For \( x > 6 \): Choose \( x = 7 \) → \( (7 - 6)(7 - 1) = 6 > 0 \)

Thus, the solution to the inequality \( (x - 6)(x - 1) \leq 0 \) is:

\[ x \in [1, 6] \]

### Final Solution

Combining both results, we find that the solution to the original inequality is:

\[ x \in [1, 6] \]