Question

Решить неравенство f(x)<0, если f(x)=(2x-3)/(x+2)

Answer 1

Решаем методом интервалов

нули: x = 1,5 и x = -2


__+____(-2)__________-______(1,5)______+______

Ответ: x ∈ (-2; 1,5)

Answer 2

To solve the inequality $f(x) < 0$, where $f(x) = \frac{{2x - 3}}{{x + 2}}$, follow these steps:

Step 1: Find the critical points of $f(x)$, where $f(x) = 0$ or $f(x)$ is undefined ($x$ that makes the denominator zero).

Step 2: Determine the sign of $f(x)$ in the intervals created by the critical points.

Step 3: Write the solution using interval notation.

Let's start with Step 1:

To find the critical points, set the numerator of $f(x)$ equal to zero:

$2x - 3 = 0$

Solve for $x$:

$2x = 3$

$x = \frac{3}{2}$

Next, set the denominator of $f(x)$ equal to zero to find any values of $x$ that make $f(x)$ undefined:

$x + 2 = 0$

$x = -2$

Step 1 results in two critical points: $x = \frac{3}{2}$ and $x = -2$.

Step 2:

Now, let's determine the sign of $f(x)$ in the intervals created by the critical points and the points that make $f(x)$ undefined:

1. For $x < -2$: Pick a test point $x = -3$ (any value less than -2), and evaluate $f(x)$:

$f(-3) = \frac{{2(-3) - 3}}{{-3 + 2}} = \frac{{-9}}{-1} = 9$

Since $f(-3)$ is positive, $f(x)$ is positive for $x < -2$.

2. For $-2 < x < \frac{3}{2}$: Pick a test point $x = 0$ (any value between -2 and $\frac{3}{2}$), and evaluate $f(x)$:

$f(0) = \frac{{2(0) - 3}}{{0 + 2}} = \frac{{-3}}{2}$

Since $f(0)$ is negative, $f(x)$ is negative for $-2 < x < \frac{3}{2}$.

3. For $x > \frac{3}{2}$: Pick a test point $x = 2$ (any value greater than $\frac{3}{2}$), and evaluate $f(x)$:

$f(2) = \frac{{2(2) - 3}}{{2 + 2}} = \frac{1}{2}$

Since $f(2)$ is positive, $f(x)$ is positive for $x > \frac{3}{2}$.

Step 3:

Write the solution using interval notation:

The solution to the inequality $f(x) < 0$ is:

$-2 < x < \frac{3}{2}$

In interval notation, this means $x$ lies between $-2$ and $\frac{3}{2}$ (exclusive).