Positive and Negative Domains: With fractions

First, we need to find the roots of the quadratic equation:

The quadratic is given by:
$y = -\frac{1}{7}x^2 - \frac{17}{7}x$

Setting $y = 0$ to find the $x$-intercepts (roots):
$-\frac{1}{7}x^2 - \frac{17}{7}x = 0$

Factor out the common factor, $-\frac{1}{7}x$:
$-\frac{1}{7}x(x + 17) = 0$

This gives the roots:
$x = 0$ and $x + 17 = 0 \rightarrow x = -17$

These roots divide the number line into intervals. We need to determine where $y > 0$. Because the coefficient of $x^2$ is negative, the parabola opens downward. The function will be positive between the roots.

Thus, we test the interval:
$(-17, 0)$

Since the parabola opens downward, the function $y > 0$ is true in the interval $-17 < x < 0$.

Therefore, the solution to the problem is $-17 < x < 0$, which corresponds to choice 3.

Let's begin by finding the roots of the quadratic function $y = -x^2 + \frac{3}{4}x - 2$. This will help us determine the intervals where the function is positive or negative.

The coefficients of the quadratic are $a = -1$, $b = \frac{3}{4}$, and $c = -2$. Applying the quadratic formula:

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

Substitute in the values:

$x = \frac{-\frac{3}{4} \pm \sqrt{\left(\frac{3}{4}\right)^2 - 4(-1)(-2)}}{2(-1)}$ $x = \frac{-\frac{3}{4} \pm \sqrt{\frac{9}{16} - 8}}{-2}$

Calculate inside the square root:

$\frac{9}{16} - 8 = \frac{9}{16} - \frac{128}{16} = \frac{-119}{16}$

The discriminant $b^2 - 4ac$ is negative ($\frac{-119}{16}$), indicating there are no real roots. Therefore, the quadratic function doesn't intersect the x-axis.

Since the parabola is downward (the coefficient of $x^2$ is negative), it is negative for all $x \in \mathbb{R}$.

We conclude that:

• The function $y$ does not have a positive domain for any real $x$.
• The function $y$ is negative for all $x$.

Therefore, the positive and negative domains, based on choices given, are:

$x > 0$: none

$x < 0$: all $x$

Let's solve the problem by following the outlined steps:

Step 1: Solve the Quadratic Equation.
First, solve the equation $\frac{1}{2}x^2 - 8 = 0$ to find the critical values:

$\frac{1}{2}x^2 - 8 = 0$

$\frac{1}{2}x^2 = 8$

$x^2 = 16$

$x = \pm 4$

Thus, the roots are $x = 4$ and $x = -4$.

Step 2: Determine Intervals and Test Sign of Function.
The roots divide the number line into three intervals: $x < -4$, $-4 < x < 4$, and $x > 4$.

• For $x < -4$, choose a test point like $x = -5$:

$y = \frac{1}{2}(-5)^2 - 8 = \frac{1}{2} \times 25 - 8 = 12.5 - 8 = 4.5$ (Positive)

• For $-4 < x < 4$, choose a test point like $x = 0$:

$y = \frac{1}{2}(0)^2 - 8 = -8$ (Negative)

• For $x > 4$, choose a test point like $x = 5$:

$y = \frac{1}{2}(5)^2 - 8 = \frac{1}{2} \times 25 - 8 = 12.5 - 8 = 4.5$ (Positive)

Conclusion:
Therefore, the function is negative in the interval where $-4 < x < 4$.

Thus, the solution for the inequality $\frac{1}{2}x^2 - 8 < 0$ is $-4 < x < 4$.

Let's solve the problem step by step:

The function given is $y = \frac{1}{5}x^2 + 1\frac{1}{3}x$. The first step is to convert the mixed number into an improper fraction:

$1\frac{1}{3} = \frac{4}{3}$, so the function becomes:

$y = \frac{1}{5}x^2 + \frac{4}{3}x$.

This can be written in standard form $ax^2 + bx + c = 0$ where $a = \frac{1}{5}$, $b = \frac{4}{3}$, and $c = 0$ (not visible, but necessary for proper representation).

Next, we use the quadratic formula to find the roots:

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

Plug in the values:

$x = \frac{-\frac{4}{3} \pm \sqrt{\left(\frac{4}{3}\right)^2 - 4 \times \frac{1}{5} \times 0}}{2 \times \frac{1}{5}}$.

Simplify where possible:

$x = \frac{-\frac{4}{3} \pm \sqrt{\frac{16}{9}}}{\frac{2}{5}}$, leading to:

$x = \frac{-\frac{4}{3} \pm \frac{4}{3}}{\frac{2}{5}}$.

This gives roots:

$x = 0$ and $x = -6\frac{2}{3}$.

The parabola opens upward (since $a = \frac{1}{5} > 0$). The quadratic function is positive between the roots and for values outside them:

This results in two intervals where $f(x) > 0$:

• $x < -6\frac{2}{3}$
• $x > 0$

Thus, the solution to the problem is:

$x > 0$ or $x < -6\frac{2}{3}$.

Let us solve the problem step by step to find: $x$ values for which f(x) < 0 .

Firstly, identify the roots of the function $y = \left(x - \frac{1}{2}\right)\left(x + 6\frac{1}{2}\right)$:

• The root from $\left(x - \frac{1}{2}\right) = 0$ is $x = \frac{1}{2}$.
• The root from $\left(x + 6\frac{1}{2}\right) = 0$ is $x = -6\frac{1}{2}$.

These roots divide the real number line into three intervals:

• $x < -6\frac{1}{2}$
• $-6\frac{1}{2} < x < \frac{1}{2}$
• $x > \frac{1}{2}$

To determine where the function is negative, evaluate the sign in each interval:

• For $x < -6\frac{1}{2}$: Both factors $(x - \frac{1}{2})$ and $(x + 6\frac{1}{2})$ are negative, so their product is positive.
• For $-6\frac{1}{2} < x < \frac{1}{2}$: $(x - \frac{1}{2})$ is negative and $(x + 6\frac{1}{2})$ is positive, thus the product is negative.
• For $x > \frac{1}{2}$: Both factors are positive, so their product is positive.

Hence, the function is negative on the interval: $-6\frac{1}{2} < x < \frac{1}{2}$.

The function given is $y = \frac{1}{3}x^2 + \frac{7}{3}x$. Our goal is to determine when this function is greater than 0.

Firstly, we set the function equal to 0 to find the critical points:

$\frac{1}{3}x^2 + \frac{7}{3}x = 0$

Factor out $\frac{1}{3}x$ from the equation:

$\frac{1}{3}x(x + 7) = 0$

This gives us two roots: $x = 0$ and $x = -7$.

Now, consider the intervals determined by these roots: $x < -7$, $-7 < x < 0$, and $x > 0$. Analyze the sign of $y$ in each interval by selecting test points.

• Interval $x < -7$: Choose $x = -8$. Substituting into the function gives $\frac{1}{3}(-8)^2 + \frac{7}{3}(-8) = \frac{64}{3} - \frac{56}{3} = \frac{8}{3} > 0$
• Interval $-7 < x < 0$: Choose $x = -1$. Substituting into the function gives $\frac{1}{3}(-1)^2 + \frac{7}{3}(-1) = \frac{1}{3} - \frac{7}{3} = -\frac{6}{3} = -2 < 0$
• Interval $x > 0$: Choose $x = 1$. Substituting into the function gives $\frac{1}{3}(1)^2 + \frac{7}{3}(1) = \frac{1}{3} + \frac{7}{3} = \frac{8}{3} > 0$

From this analysis, the function $y$ is positive when $x < -7$ or $x > 0$. Thus, the solution is:

The function $f(x)$ is positive for $x > 0$ or $x < -7$.

The function $y = \frac{1}{2}x^2 - 2$ is quadratic. To find where $f(x) < 0$, identify where $f(x) = 0$ by setting the equation equal to zero and solving for $x$.

The equation is $\frac{1}{2}x^2 - 2 = 0$.

Multiply through by 2 to eliminate the fraction:
$x^2 - 4 = 0$.

Set the equation as $(x - 2)(x + 2) = 0$ to find the roots.

Solving gives roots $x = 2$ and $x = -2$.

The function $f(x)$ will change signs at these roots $x = -2$ and $x = 2$.

Check intervals determined by the roots to find where $f(x) < 0$:

• For $-2 < x < 2$: Evaluate one test point like $x = 0$ in $f(x)$:
  $f(0) = \frac{1}{2}(0)^2 - 2 = -2$, indicating $f(x) < 0$.
• For intervals $x < -2$ or $x > 2$: Select $x = -3$ and $x = 3$ respectively, both yield positive results, therefore $f(x) > 0$.

Ultimately, the function $f(x)$ is negative only on the interval $-2 < x < 2$.

Therefore, the values of $x$ for which $f(x) < 0$ are such that $-2 < x < 2$.

The function $y = \left(\frac{1}{3}x + \frac{1}{6}\right)\left(-x - 4\frac{1}{5}\right)$ requires us to analyze the sign of the product for various $x$ values.

First, we must find the zeros of each factor:

• The zero of $\frac{1}{3}x + \frac{1}{6}$ is found by solving $\frac{1}{3}x + \frac{1}{6} = 0$:
  Subtract $\frac{1}{6}$ to get:
  $\frac{1}{3}x = -\frac{1}{6}$
  Multiply both sides by 3:
  $x = -\frac{1}{2}$.
• The zero of $-x - 4\frac{1}{5}$ is found by solving $-x - 4\frac{1}{5} = 0$:
  Add $4\frac{1}{5}$ to get:
  $-x = 4\frac{1}{5}$
  Multiply by $-1$ to find:
  $x = -4\frac{1}{5}$.

Next, we identify the intervals defined by these zeros: $x < -4\frac{1}{5}$, $-4\frac{1}{5} < x < -\frac{1}{2}$, and $x > -\frac{1}{2}$.

We will determine the sign of the function in each interval:

• In $x < -4\frac{1}{5}$:
  - $\frac{1}{3}x + \frac{1}{6}$ and $-x - 4\frac{1}{5}$ are both negative (since both points are below their respective roots), resulting in a positive product.
• In $-4\frac{1}{5} < x < -\frac{1}{2}$:
  - $\frac{1}{3}x + \frac{1}{6}$ is negative and $-x - 4\frac{1}{5}$ is positive, resulting in a negative product.
• In $x > -\frac{1}{2}$:
  - $\frac{1}{3}x + \frac{1}{6}$ and $-x - 4\frac{1}{5}$ are both positive, resulting in a positive product.

The function is negative in the interval $-4\frac{1}{5} < x < -\frac{1}{2}$. Thus, the correct answer corresponding to where the function is negative is the complementary intervals $x > -\frac{1}{2}$ or $x < -4\frac{1}{5}$, which matches choice 2.

Therefore, the solution is $x > -\frac{1}{2}$ or $x < -4\frac{1}{5}$.

The function we are given is $y = x^2 - \frac{4}{9}$. This is a quadratic function.

To find where $f(x) > 0$, we first need to determine where the function equals zero and changes sign. This involves solving the equation:

$x^2 - \frac{4}{9} = 0$

Rearranging gives:

$x^2 = \frac{4}{9}$

Taking the square root of both sides, we find:

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

These are the points where the function changes signs. The parabola represented by this quadratic function opens upwards (since the coefficient of $x^2$ is positive and equal to 1), indicating that it is positive outside the interval between these roots and negative inside:

• The intervals of interest are $x < -\frac{2}{3}$, $-\frac{2}{3} < x < \frac{2}{3}$, and $x > \frac{2}{3}$.
• In the interval $x < -\frac{2}{3}$, $x^2 - \frac{4}{9} > 0$ because outside the roots, the parabola is above the x-axis.
• In the interval $-\frac{2}{3} < x < \frac{2}{3}$, $x^2 - \frac{4}{9} < 0$ because it is between the roots where the parabola lies below the x-axis.
• In the interval $x > \frac{2}{3}$, $x^2 - \frac{4}{9} > 0$ for the same reason as $x < -\frac{2}{3}$.

Therefore, the function $f(x) > 0$ when $x < -\frac{2}{3}$ or $x > \frac{2}{3}$.

Considering the choices provided, the correct answer that satisfies $f(x) > 0$ is choice 3: $x > \frac{2}{3}$ or $x < -\frac{2}{3}$.

The given function is $y = -\frac{1}{2}x^2 + 2\frac{25}{72}$. We need to find when this function is equal to zero to determine the positive and negative domains.

First, identify the coefficients from the function:

• $a = -\frac{1}{2}$, $b = 0$, $c = 2\frac{25}{72}$ which we convert to improper fraction: $\frac{169}{72}$.

We then use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to find the roots.

Substituting in our values:

$x = \frac{-0 \pm \sqrt{0 - 4\left(-\frac{1}{2}\right)\left(\frac{169}{72}\right)}}{-1}$.

The discriminant calculation is as follows:

$4 \cdot \frac{1}{2} \cdot \frac{169}{72} = \frac{169}{36}$.

So our roots are:

$x = \pm \sqrt{\frac{169}{36}} = \pm \frac{13}{6} = \pm 2\frac{1}{6}$.

The roots are $x = 2\frac{1}{6}$ and $x = -2\frac{1}{6}$. These divide the x-axis into intervals to be tested.

- When $x < -2\frac{1}{6}$, test point $x = -3$:
$y = -\frac{1}{2}(3)^2 + \frac{169}{72} < 0$: Negative.
Hence, $x < -2\frac{1}{6}$ gives negative values.

- When $-2\frac{1}{6} < x < 2\frac{1}{6}$, test $x = 0$:
$y = -\frac{1}{2}(0)^2 + \frac{169}{72} > 0$: Positive.
Hence, $-2\frac{1}{6} < x < 2\frac{1}{6}$ gives positive values.

- When $x > 2\frac{1}{6}$, test point $x = 3$:
$y = -\frac{1}{2}(3)^2 + \frac{169}{72} < 0$: Negative.
Hence, $x > 2\frac{1}{6}$ gives negative values.

Therefore, the positive domain is $x > 0 : -2\frac{1}{6} < x < 2\frac{1}{6}$ and the negative domain is $x < 0 : x < -2\frac{1}{6}$ or $x > 2\frac{1}{6}$.

In comparing to the provided choices, the correct choice is Choice 2: $x > 2\frac{1}{6}$ or $x < 0 : x < -2\frac{1}{6}$

$x > 0 : - 2\frac{1}{6} < x < 2\frac{1}{6}$

The quadratic function is $y = -x^2 - \frac{4}{3}$. This function graphs as a parabola opening downwards.

Let's analyze the sign of the function:

• Since the parabola opens downwards (negative coefficient of $x^2$), $y$ takes on non-positive values. No solution exists for when $y > 0$.
• For $y = 0$, solving gives: $-x^2 - \frac{4}{3} = 0$ which means $-x^2 = \frac{4}{3}$. This equation cannot be true for any real $x$ because squares are non-negative and cannot yield negative values.
• Therefore, the function does not cross or even touch the $x$-axis.
• This means the range of the function simply yields negative values for any $x$.

Let's review some regions:

• Negative Domain, $x < 0$: Since the function always remains negative or zero, for negative $x$, $\,$ the domain technically spans all real numbers, but function values will be less than 0 (negative).
• Positive Domain, $x > 0$: None as $y > 0$ does not happen for any real $x$ due to the nature of the parabola being downward opening and completely below the $x$-axis.

The correct choice identifying domains is: $x < 0 :$ all $x$

$x > 0 :$ none.

Therefore, the positive and negative domains of the function are:

x < 0 : 1 < x < 1.5

x > 1.5 or x > 0 : x < 1

To determine the positive and negative domains of the function $y = -x^2 + 5x + 6$, we first find the roots of the equation by solving:

$-x^2 + 5x + 6 = 0$.

We use the quadratic formula:

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

Here, $a = -1$, $b = 5$, and $c = 6$. Substituting these values, we find:

$x = \frac{-5 \pm \sqrt{5^2 - 4(-1)(6)}}{2(-1)} = \frac{-5 \pm \sqrt{25 + 24}}{-2} = \frac{-5 \pm \sqrt{49}}{-2}$.

$x = \frac{-5 \pm 7}{-2}$.

Solving the two scenarios regarding the $\pm$ gives $x = \frac{2}{-2} = -1$ and $x = \frac{-12}{-2} = 6$.

This means the roots are $x = -1$ and $x = 6$.

We now test the intervals defined by these roots: $(-\infty, -1)$, $(-1, 6)$, and $(6, \infty)$.

- For $x < -1$: pick $x = -2$. Substitute into the function:

$y = -(-2)^2 + 5(-2) + 6 = -4 - 10 + 6 = -8$ (negative).

- For $-1 < x < 6$: pick $x = 0$. Substitute:

$y = -(0)^2 + 5(0) + 6 = 6$ (positive).

- For $x > 6$: pick $x = 7$. Substitute:

$y = -(7)^2 + 5(7) + 6 = -49 + 35 + 6 = -8$ (negative).

Thus, the function is positive in the interval $-1 < x < 6$ and negative in the intervals $x < -1$ and $x > 6$.

Therefore, the solution to the problem is:

x > 0 : -1 < x < 6

x > 6 or x < 0 : x < -1

To determine the positive and negative domains of the quadratic function $y = 3x^2 + 7x + 2$, we will first find its roots using the quadratic formula.

The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

For this function, $a = 3$, $b = 7$, and $c = 2$.

First, calculate the discriminant $\Delta$:

$\Delta = b^2 - 4ac = 7^2 - 4 \cdot 3 \cdot 2 = 49 - 24 = 25$.

Since the discriminant is positive, the function has two distinct real roots.

Now, calculate the roots:

$x = \frac{-b \pm \sqrt{\Delta}}{2a}$.

$x = \frac{-7 \pm \sqrt{25}}{6} = \frac{-7 \pm 5}{6}$.

This results in:

• Root 1: $x = \frac{-7 + 5}{6} = \frac{-2}{6} = -\frac{1}{3}$
• Root 2: $x = \frac{-7 - 5}{6} = \frac{-12}{6} = -2$

The roots are $x = -2$ and $x = -\frac{1}{3}$, dividing the x-axis into three intervals: $x < -2$, $-2 < x < -\frac{1}{3}$, and $x > -\frac{1}{3}$.

For $x \to -\infty$, the quadratic $y = 3x^2 + 7x + 2$ is positive, as the leading coefficient $a = 3$ is positive, indicating the parabola opens upwards.

Evaluate the sign of $y$ within the intervals:

• For $x < -2$, pick $x = -3$: $y = 3(-3)^2 + 7(-3) + 2 = 27 - 21 + 2 = 8$. Hence, $y > 0$.
• For $-2 < x < -\frac{1}{3}$, pick $x = -1$: $y = 3(-1)^2 + 7(-1) + 2 = 3 - 7 + 2 = -2$. Hence, $y < 0$.
• For $x > -\frac{1}{3}$, pick $x = 0$: $y = 3(0)^2 + 7(0) + 2 = 2$. Hence, $y > 0$.

Therefore, the positive domain of the function is $x < -2$ and $x > -\frac{1}{3}$, and the negative domain is $-2 < x < -\frac{1}{3}$.

Thus, the solution matches the given correct answer:

$x < 0 : -2 < x < -\frac{1}{3}$

$x > -\frac{1}{3}$ or $x > 0 : x < -2$

To determine the positive and negative domains of the quadratic function $y = \frac{1}{2}x^2 + \frac{3}{4}x + \frac{5}{6}$, we start by considering the possibility of real roots using the discriminant.

The discriminant $\Delta$ is given by:

$\Delta = b^2 - 4ac = \left(\frac{3}{4}\right)^2 - 4\left(\frac{1}{2}\right)\left(\frac{5}{6}\right)$

Calculating gives:

$\Delta = \frac{9}{16} - \frac{20}{12} = \frac{9}{16} - \frac{5}{3} = \frac{9}{16} - \frac{80}{48}$

Convert $\frac{9}{16}$ to a common denominator:

$\Delta = \frac{27}{48} - \frac{80}{48} = -\frac{53}{48}$

The discriminant $\Delta$ is negative, indicating that this quadratic equation has no real roots.

Since the coefficient $a = \frac{1}{2}$ is positive and there are no real roots, the parabola opens upwards and never crosses the x-axis.

This means that the function is always positive for all $x$.

Thus, the positive domain is all $x$, and there is no negative domain.

Therefore, the correct choice is:

$x > 0 :$ for all $x$

$x < 0 :$ none

To determine the positive and negative domains of the quadratic function $y = \frac{1}{3}x^2 + x + 2$, we will follow these general steps:

• Find the roots of the equation $\frac{1}{3}x^2 + x + 2 = 0$.
• Analyze the sign of the quadratic between and beyond these roots.

First, let's identify the roots of the quadratic function:

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = \frac{1}{3}$, $b = 1$, and $c = 2$, the discriminant $\Delta = b^2 - 4ac = 1^2 - 4 \cdot \frac{1}{3} \cdot 2 = 1 - \frac{8}{3} = \frac{-5}{3}$.

Since the discriminant is negative, the quadratic equation $\frac{1}{3}x^2 + x + 2 = 0$ has no real roots.

Given that the coefficient of $x^2$ (i.e., $\frac{1}{3}$) is positive, the parabola opens upwards, meaning the entire function is greater than zero for all real values of $x$.

Therefore, the positive domain is all real numbers, and there is no negative domain.

Therefore, the solution is: $x > 0$: for all $x$; $x < 0$: none.

To determine when $f(x) = \frac{1}{4}x^2 - 3\frac{1}{2}x < 0$, follow these steps:

Step 1: Find the roots of the quadratic equation.

The function can be rewritten as $y = \frac{1}{4}x^2 - \frac{7}{2}x$. Set this equal to zero to find the roots:

$\frac{1}{4}x^2 - \frac{7}{2}x = 0$

Factor out $x$: $x\left(\frac{1}{4}x - \frac{7}{2}\right) = 0$

So, $x = 0$ or $\frac{1}{4}x = \frac{7}{2}$. Solve the second equation:

$x = \frac{7}{2} \times 4 = 14$

Step 2: Analyze the intervals around the roots.

The roots are $x = 0$ and $x = 14$. These divide the number line into three intervals: $x < 0$, $0 < x < 14$, and $x > 14$.

Step 3: Perform a sign test in each interval.

• Test for $x < 0$: Choose $x = -1$. The value of the function $f(-1) = \frac{1}{4}(-1)^2 - \frac{7}{2}(-1) = \frac{1}{4} + \frac{7}{2} > 0$.
• Test for $0 < x < 14$: Choose $x = 7$. The value of the function $f(7) = \frac{1}{4}(7)^2 - \frac{7}{2}(7) = \frac{49}{4} - \frac{49}{2} = \frac{49}{4} - \frac{98}{4} = -\frac{49}{4} < 0$.
• Test for $x > 14$: Choose $x = 15$. The value of the function $f(15) = \frac{1}{4}(15)^2 - \frac{7}{2}(15) = \frac{225}{4} - \frac{105}{2} = \frac{225}{4} - \frac{210}{4} = \frac{15}{4} > 0$.

Conclusion: The quadratic $\frac{1}{4}x^2 - \frac{7}{2}x$ is less than zero for $0 < x < 14$.

Therefore, the solution to the problem is $0 < x < 14$.

To determine when the quadratic function $y = (x - 4.4)(x - 2.3)$ is negative, we need to analyze the sign of the product across the different intervals defined by its roots.

• Step 1: Identify the roots of the function. The roots occur when each factor equals zero, which are $x = 2.3$ and $x = 4.4$.
• Step 2: Divide the x-axis into intervals based on these roots: $x < 2.3$, $2.3 < x < 4.4$, and $x > 4.4$.
• Step 3: Test a value from each interval:
  • For $x < 2.3$, try $x = 2$: $y = (2 - 4.4)(2 - 2.3) = (-2.4)(-0.3) = 0.72$, so the product is positive.
  • For $2.3 < x < 4.4$, try $x = 3$: $y = (3 - 4.4)(3 - 2.3) = (-1.4)(0.7) = -0.98$, so the product is negative.
  • For $x > 4.4$, try $x = 5$: $y = (5 - 4.4)(5 - 2.3) = (0.6)(2.7) = 1.62$, so the product is positive.

From this analysis, we see that the quadratic function is negative for values of $x$ in the interval $2.3 < x < 4.4$. This is the range where the function changes sign from positive to negative back to positive.

Therefore, the correct answer is $2.3 < x < 4.4$.