Question

(x + 3)/(2 + x ^ 2) < 3

Answer 1

Ответ:

Объяснение:

!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 2

To solve the inequality $\frac{x + 3}{2 + x^2} < 3$, we'll first find the critical points where the expression is equal to 3, and then determine the intervals where it is less than 3.

Step 1: Find the critical points (where the expression is equal to 3). $\frac{x + 3}{2 + x^2} = 3$

Multiply both sides by $(2 + x^2)$ to eliminate the denominator:

$x + 3 = 3(2 + x^2)$

Now, distribute the 3 on the right side:

$x + 3 = 6 + 3x^2$

Rearrange the terms to get the quadratic equation:

$3x^2 - x - 3 = 0$

Step 2: Solve the quadratic equation for $x$. To solve this quadratic equation, you can use the quadratic formula:

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

In this case, $a = 3$, $b = -1$, and $c = -3$.

$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(3)(-3)}}{2(3)}$

Simplify:

$x = \frac{1 \pm \sqrt{1 + 36}}{6}$

$x = \frac{1 \pm \sqrt{37}}{6}$

So, the critical points are $x = \frac{1 + \sqrt{37}}{6}$ and $x = \frac{1 - \sqrt{37}}{6}$.

Step 3: Determine the intervals where the expression is less than 3. We have three critical points: $\frac{1 + \sqrt{37}}{6}$, $\frac{1 - \sqrt{37}}{6}$, and the endpoints of the real number line.

Now, we'll test the inequality $\frac{x + 3}{2 + x^2} < 3$ in each of these intervals.

1. Interval $(-\infty, \frac{1 - \sqrt{37}}{6})$: Choose a test point $x < \frac{1 - \sqrt{37}}{6}$, e.g., $x = 0$: $\frac{0 + 3}{2 + 0^2} < 3$ $\frac{3}{2} < 3$ This is true.

2. Interval $\left(\frac{1 - \sqrt{37}}{6}, \frac{1 + \sqrt{37}}{6}\right)$: Choose a test point $\frac{1 - \sqrt{37}}{6} < x < \frac{1 + \sqrt{37}}{6}$