Question

При каких значениях x f(x)<0, если f(x)=3-2х/4х+1

Answer 1

$f(x)=\dfrac{3-2x}{4x+1}$

Область определения функции: функция существует, когда знаменатель дроби не обращается к нулю

$4x+1\ne 0\\ x\ne -0.25$

Приравниваем теперь функцию к нулю, т.е. f(x) = 0

$\dfrac{3-2x}{4x+1}=0~~~\Rightarrow~~~ 3-2x=0~~~\Rightarrow~~~ x=1.5$

Ответ: при $x \in (-\infty;-0.25)\cup(1.5;+\infty)$

Answer 2

х ∈ (-0.3; -0.25)

(фото)

Answer 3

To determine the values of x for which f(x) < 0, we need to solve the inequality f(x) < 0. Let's work through the steps:

Given: f(x) = (3 - 2x) / (4x + 1)

To solve f(x) < 0, we can start by finding the critical points where f(x) changes sign. In this case, the critical points occur when f(x) is equal to zero or undefined. So, let's set the numerator and denominator equal to zero:

3 - 2x = 0 (numerator equal to zero) 4x + 1 = 0 (denominator equal to zero)

Solving these equations, we find:

3 - 2x = 0 2x = 3 x = 3/2

4x + 1 = 0 4x = -1 x = -1/4

So, the critical points are x = 3/2 and x = -1/4.

Now, let's test the intervals defined by these critical points to determine when f(x) < 0. We can choose a test point from each interval and substitute it into f(x) to determine the sign of f(x) in that interval.

1. Interval (-∞, -1/4): Let's choose x = -1 as our test point. Substituting x = -1 into f(x), we get: f(-1) = (3 - 2(-1)) / (4(-1) + 1) = (3 + 2) / (-4 + 1) = 5 / -3 Since 5 / -3 is negative, f(x) < 0 in the interval (-∞, -1/4).

2. Interval (-1/4, 3/2): Let's choose x = 0 as our test point. Substituting x = 0 into f(x), we get: f(0) = (3 - 2(0)) / (4(0) + 1) = 3 / 1 = 3 Since 3 is positive, f(x) > 0 in the interval (-1/4, 3/2).

3. Interval (3/2, +∞): Let's choose x = 2 as our test point. Substituting x = 2 into f(x), we get: f(2) = (3 - 2(2)) / (4(2) + 1) = (3 - 4) / (8 + 1) = -1 / 9 Since -1 / 9 is negative, f(x) < 0 in the interval (3/2, +∞).

To summarize: f(x) < 0 in the intervals (-∞, -1/4) and (3/2, +∞).

Note: The critical points and intervals are determined by analyzing the numerator and denominator separately and considering the sign changes.