Question

(x-2)^2 (x-4) /x+2<0

Answer 1

Решение смотри на фотографии

Answer 2

To solve the inequality $\frac{(x-2)^2 (x-4)}{x+2} < 0$, we need to find the values of $x$ that satisfy the inequality. To do this, we can follow these steps:

Step 1: Find the critical points by setting the numerator and denominator equal to zero and solving for $x$. Step 2: Determine the sign of the expression $\frac{(x-2)^2 (x-4)}{x+2}$ in each interval formed by the critical points. Step 3: Identify the intervals where the expression is negative (less than zero).

Let's start with step 1:

Step 1: Find the critical points: The critical points occur when the numerator or the denominator is equal to zero. Numerator: $(x - 2)^2 (x - 4) = 0$ This gives us two critical points: $x = 2$ and $x = 4$.

Denominator: $x + 2 = 0$ This gives us another critical point: $x = -2$.

Step 2: Determine the sign of the expression in each interval: We now have four intervals: $(- \infty, -2)$, $(-2, 2)$, $(2, 4)$, and $(4, \infty)$. We will check the sign of the expression in each interval.

For $x < -2$: Choose $x = -3$ (a value less than -2) $\frac{(-3-2)^2 (-3-4)}{-3+2} = \frac{25}{-1} = -25$, which is negative.

For $-2 < x < 2$: Choose $x = 0$ (a value between -2 and 2) $\frac{(0-2)^2 (0-4)}{0+2} = \frac{16}{2} = 8$, which is positive.

For $2 < x < 4$: Choose $x = 3$ (a value between 2 and 4) $\frac{(3-2)^2 (3-4)}{3+2} = \frac{-1}{5}$, which is negative.

For $x > 4$: Choose $x = 5$ (a value greater than 4) $\frac{(5-2)^2 (5-4)}{5+2} = \frac{9}{7}$, which is positive.

Step 3: Identify the intervals where the expression is negative: The expression is negative in the intervals $(- \infty, -2)$ and $(2, 4)$.

So, the solution to the inequality $\frac{(x-2)^2 (x-4)}{x+2} < 0$ is $x \in (-\infty, -2) \cup (2, 4)$.