Question

1<(3х^2-6х+12)/(х^2+2)<3

Answer 1

Решение на фотке
Верю что понятно

Answer 2

To solve the inequality $\frac{3x^2 - 6x + 12}{x^2 + 2}$ for $x$, we can follow these steps:

Step 1: Find the domain of the expression Since the denominator $x^2 + 2$ cannot be equal to zero (as division by zero is undefined), we need to find the values of $x$ for which $x^2 + 2 = 0$. However, there are no real solutions for this equation because $x^2 + 2 = 0$ leads to $x^2 = -2$, which has no real solutions. Thus, the expression is defined for all real values of $x$.

Step 2: Determine the critical points of the expression Critical points occur where the numerator or denominator equals zero. However, since there are no real solutions for $x^2 + 2 = 0$, there are no critical points in this case.

Step 3: Analyze the sign of the expression in different intervals Since there are no critical points, we can choose test points in different intervals to determine the sign of the expression in those intervals.

Let's choose $x = -1$, $x = 0$, and $x = 1$ as test points, which divide the real number line into four intervals: $(-∞, -1)$, $(-1, 0)$, $(0, 1)$, and $(1, +∞)$.

For $x = -1$: $\frac{3(-1)^2 - 6(-1) + 12}{(-1)^2 + 2} = \frac{3 + 6 + 12}{3} = \frac{21}{3} = 7$. Thus, the expression is positive in the interval $(-∞, -1)$.

For $x = 0$: $\frac{3(0)^2 - 6(0) + 12}{(0)^2 + 2} = \frac{12}{2} = 6$. Thus, the expression is positive in the interval $(-1, 0)$.

For $x = 1$: $\frac{3(1)^2 - 6(1) + 12}{(1)^2 + 2} = \frac{3 - 6 + 12}{3} = \frac{9}{3} = 3$. Thus, the expression is positive in the interval $(0, 1)$.

Step 4: Solve the inequality The inequality is $1 < \frac{3x^2 - 6x + 12}{x^2 + 2} < 3$.

Since the expression is positive in all intervals, we don't need to worry about the direction of the inequality signs.

The solution is all real values of $x$ because the expression $\frac{3x^2 - 6x + 12}{x^2 + 2}$ is positive for all real values of $x$, and thus, the inequality is satisfied for all $x$.