Standard Representation: With fractions

To solve this problem, we'll find the intervals where the given quadratic function $y = -2x^2 + 7x - 3$ is greater than zero (positive) and less than zero (negative).

Step 1: Find the roots of the quadratic function.

The general form of the quadratic equation is $ax^2 + bx + c = 0$. Here, $a = -2$, $b = 7$, and $c = -3$.

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we calculate the roots.

First, calculate the discriminant:

$b^2 - 4ac = 7^2 - 4 \times (-2) \times (-3) = 49 - 24 = 25$.

Thus, the roots are:

$x = \frac{-7 \pm \sqrt{25}}{2 \times (-2)} = \frac{-7 \pm 5}{-4}$.

Calculating for the two roots:

• $x_1 = \frac{-7 + 5}{-4} = \frac{-2}{-4} = \frac{1}{2}$
• $x_2 = \frac{-7 - 5}{-4} = \frac{-12}{-4} = 3$

The roots are $x = \frac{1}{2}$ and $x = 3$.

Step 2: Determine the sign of the function in each interval.

The function is defined as:

$(-\infty, \frac{1}{2}), \left(\frac{1}{2}, 3\right), (3, \infty)$.

Test each interval to determine where the function is positive or negative:

• For $x < \frac{1}{2}$, choose $x = 0$:
  $y = -2(0)^2 + 7(0) - 3 = -3$ (negative)
• For $\frac{1}{2} < x < 3$, choose $x = 1$:
  $y = -2(1)^2 + 7(1) - 3 = 2$ (positive)
• For $x > 3$, choose $x = 4$:
  $y = -2(4)^2 + 7(4) - 3 = -3$ (negative)

Conclusion: The positive domain is $\frac{1}{2} < x < 3$, and the negative domain is $x < \frac{1}{2}$ or $x > 3$.

Therefore, the correct option is:

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

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

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 $x$

To find the positive and negative domains of the function $y = -x^2 + \frac{3}{2}x - \frac{21}{4}$, we first determine the roots of the equation:

Set $y = 0$, giving us:

$-x^2 + \frac{3}{2}x - \frac{21}{4} = 0$.

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = -1$, $b = \frac{3}{2}$, and $c = -\frac{21}{4}$, we calculate:

• First, compute the discriminant: $b^2 - 4ac = \left(\frac{3}{2}\right)^2 - 4(-1)\left(-\frac{21}{4}\right) = \frac{9}{4} - \frac{21}{1} = \frac{9}{4} - \frac{84}{4} = -\frac{75}{4}$.
• Since the discriminant is negative, the quadratic has no real roots.

This implies that the parabola does not intersect the $x$-axis and since the quadratic coefficient $a = -1$ is negative, the parabola opens downwards.

Thus, the function is always negative for all $x$. Therefore, the positive domain is empty, and the negative domain is the entire set of real numbers.

Conclusion: The solution to the problem is as follows:

$x < 0$: for all $x$

$x > 0$: none

To solve this problem, we'll follow these steps:

• Step 1: Identify the roots of the quadratic function.
• Step 2: Use the roots to determine intervals.
• Step 3: Test each interval to determine if the function is positive or negative.

Now, let's work through each step:
Step 1: The quadratic function is $y = -\frac{1}{3}x^2 + \frac{2}{3}x - \frac{1}{3}$. We apply the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to find the roots, where $a = -\frac{1}{3}$, $b = \frac{2}{3}$, and $c = -\frac{1}{3}$.

Calculating the discriminant $b^2 - 4ac$:

$\left(\frac{2}{3}\right)^2 - 4\left(-\frac{1}{3}\right)\left(-\frac{1}{3}\right) = \frac{4}{9} - \frac{4}{9} = 0$.

The discriminant is zero, indicating a repeated root at:

$x = \frac{-\frac{2}{3}}{2\left(-\frac{1}{3}\right)} = 1$.

Step 2: The repeated root is $x = 1$. For $x < 1$ and $x > 1$, evaluate the sign of the function.

Step 3: Testing intervals:

• For $x = 0$ (as a test point for $x < 1$), substitute into the original function:

$y = -\frac{1}{3}(0)^2 + \frac{2}{3}(0) - \frac{1}{3} = -\frac{1}{3}$. The function is negative.

• For $x > 1$, any positive $x$ will substitute into the function and confirm it remains negative, as the parabola opens downwards and cannot turn positive again.

Therefore, the solution to the problem is:

$x < 0 : x \ne 1$, and $x > 0$: none.

To find the positive and negative domains of the function $y = -\frac{1}{2}x^2 + \frac{1}{3}x - \frac{1}{4}$, we must determine where the function is above or below the x-axis.

Step 1: Find the roots of the quadratic equation. This requires solving:

$-\frac{1}{2}x^2 + \frac{1}{3}x - \frac{1}{4} = 0$

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, with $a = -\frac{1}{2}$, $b = \frac{1}{3}$, and $c = -\frac{1}{4}$, we calculate:

• Discriminant: $b^2 - 4ac = \left(\frac{1}{3}\right)^2 - 4\left(-\frac{1}{2}\right)\left(-\frac{1}{4}\right) = \frac{1}{9} - \frac{1}{2} = \frac{1}{9} - \frac{4}{8} = \frac{1}{9} - \frac{9}{18} = \frac{1 - 9}{18} = -\frac{8}{18} = -\frac{4}{9}$

The discriminant is negative, indicating no real roots.

Step 2: Analyze the parabola's orientation. Because the leading term is negative, the parabola opens downwards. With no x-intercepts, this implies the entire graph is below the x-axis.

Therefore, the function is negative for all x-values. In the context of positive and negative domains:

$x > 0 :$ none, as the function doesn't cross the x-axis in positive domain.

$x < 0 :$ all $x$, as the function is always negative.

To solve this problem, we'll follow these steps:

• Step 1: Identify roots of the quadratic equation.
• Step 2: Determine the intervals created by these roots.
• Step 3: Test each interval to see where the function is positive or negative.

Now, let's work through each step:

Step 1: We need to find the roots of the equation $-\frac{1}{3}x^2 + 2x - 4 = 0$. Using the quadratic formula:
For $ax^2 + bx + c = 0$, the roots are given by:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
Here, $a = -\frac{1}{3}$, $b = 2$, and $c = -4$.

Step 2: Calculate the discriminant:
$b^2 - 4ac = 2^2 - 4(-\frac{1}{3})(-4) = 4 - \frac{16}{3} = \frac{12}{3} - \frac{16}{3} = -\frac{4}{3}$.

Since the discriminant is negative, the quadratic does not have real roots. Therefore, the function does not cross the x-axis and remains entirely above or below the x-axis.

Step 3: Analyze the leading coefficient. The quadratic function opens downwards because the leading coefficient $a = -\frac{1}{3}$ is negative. Therefore, since there are no x-intercepts, the function is negative for all $x$.

Thus, we find that:
- The positive domain of $y$ is: none.
- The negative domain of $y$ is: for all $x$.

Therefore, the solution to the problem is:

$x < 0 :$ for all $x$

$x > 0 :$ none

To solve the given problem, we will perform the following steps:

• Calculate the roots of the quadratic equation $y = -2x^2 + 3x + 2$ using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Given $a = -2$, $b = 3$, and $c = 2$, use the formula:

$x = \frac{-3 \pm \sqrt{3^2 - 4(-2)(2)}}{2(-2)} = \frac{-3 \pm \sqrt{9 + 16}}{-4} = \frac{-3 \pm \sqrt{25}}{-4}$

• The roots are $x = \frac{-3 + 5}{-4} = -\frac{1}{2}$ and $x = \frac{-3 - 5}{-4} = 2$.
• The roots $-\frac{1}{2}$ and $2$ divide the x-axis into three intervals: $(-\infty, -\frac{1}{2})$, $(-\frac{1}{2}, 2)$, and $(2, \infty)$.
• Test a point from each interval in the function to determine the sign of the function in those intervals:

Choose $x = -1$ for interval $(- \infty, -\frac{1}{2})$:
Substitute into the function: $y = -2(-1)^2 + 3(-1) + 2 = -2 - 3 + 2 = -3$ (negative).

Choose $x = 0$ for interval $(- \frac{1}{2}, 2)$:
Substitute into the function: $y = -2(0)^2 + 3(0) + 2 = 2$ (positive).

Choose $x = 3$ for interval $(2, \infty)$:
Substitute into the function: $y = -2(3)^2 + 3(3) + 2 = -18 + 9 + 2 = -7$ (negative).

Therefore, the positive domain where the function is positive is $-\frac{1}{2} < x < 2$, and the negative domains are $x < -\frac{1}{2}$ or $x > 2$.

The solution to the problem is:

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

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

To find the domains where the function is positive and negative, let's follow these steps:

• Identify the quadratic function given: $y = -x^2 + \frac{1}{2}x - 3$.
• Use the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, to find the roots.

Substitute $a = -1$, $b = \frac{1}{2}$, and $c = -3$ into the quadratic formula:

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

$x = \frac{-\frac{1}{2} \pm \sqrt{\frac{1}{4} - 12}}{-2}$.

$x = \frac{-\frac{1}{2} \pm \sqrt{-\frac{47}{4}}}{-2}$.

This results in a negative discriminant ($-\frac{47}{4}$), meaning there are no real roots.

Since there are no real roots, the function does not cross the x-axis, and given the parabola opens downwards ($a < 0$), the entire curve lies below the x-axis.

Therefore, the function is negative for all $x$.

This means:
For $x < 0$: the function is negative for all $x$.
For $x > 0$: there are no positive intervals as the function is negative everywhere.

Thus, the solution indicates that the function is always negative, confirming the negative domain spans all real numbers, and the positive domain is nonexistent.

The correct choice aligning with this result is Choice 2: $x < 0$: for all $x$, and $x > 0$: none.

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 solve this problem, we need to analyze the function $y = \frac{1}{3}x^2 + \frac{1}{2}x + \frac{2}{3}$ to determine where it is positive or negative. This function is quadratic, so it is a parabola. Let us find the domain of positive and negative values.

First, let's determine where the function is zero by finding its roots. This involves using the quadratic formula:

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

Here, we have $a = \frac{1}{3}$, $b = \frac{1}{2}$, and $c = \frac{2}{3}$. The discriminant is calculated as follows:

$b^2 - 4ac = \left(\frac{1}{2}\right)^2 - 4 \cdot \frac{1}{3} \cdot \frac{2}{3} = \frac{1}{4} - \frac{8}{9}$

Calculating gives:

$\frac{1}{4} \approx 0.25$ and $\frac{8}{9} \approx 0.888$ which results in:

$0.25 - 0.888 = -0.638$

Since the discriminant is negative, there are no real roots. As the parabola opens upwards (since $a = \frac{1}{3} > 0$), the function never crosses the x-axis. Therefore, the function remains positive for all real values of $x$ and is never negative.

Thus, the positive domain consists of all real numbers, while there is no negative domain:

$x > 0$: \) for all $x$

$x < 0$: \) none

To determine the positive and negative domains of $y = 25x^2 + 20x + 4$, we follow these steps:

• Step 1: Calculate the discriminant $\Delta = b^2 - 4ac = 20^2 - 4 \cdot 25 \cdot 4 = 400 - 400 = 0$.
• Step 2: Since the discriminant is zero, the quadratic has one distinct real root, given by $x = \frac{-b}{2a} = \frac{-20}{2 \cdot 25} = -\frac{2}{5}$.
• Step 3: Analyze the sign of $y = 25x^2 + 20x + 4$.

Since the discriminant is zero, the quadratic equation reaches zero only at $x = -\frac{2}{5}$, and is symmetrical around this point. Therefore:

• For $x > 0$, $y$ is positive except where $x = -\frac{2}{5}$.
• For $x < 0$, there are no points where the graph is negative since the parabola opens upwards and does not cross below the x-axis after this root.

Thus, interpreting the domains:

The positive domain for $x > 0$ is all $x$ except $x = -\frac{2}{5}$. For $x < 0$, there is no negative domain because the graph does not descend below the x-axis.

Therefore, the solution to the problem is:

$x > 0 :x\ne\frac{2}{5}$

$x < 0 :$ 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 solve this problem, we need to find when the quadratic function changes from positive to negative and vice versa.

• Step 1: Find the roots using the quadratic formula. For $y = 2x^2 + 7x - 9$, identify $a = 2$, $b = 7$, and $c = -9$.
• Step 2: Apply the quadratic formula:
  $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
• Step 3: Calculate:
  $x = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 2 \cdot (-9)}}{2 \cdot 2}$
  $x = \frac{-7 \pm \sqrt{49 + 72}}{4}$
  $x = \frac{-7 \pm \sqrt{121}}{4}$
  $x = \frac{-7 \pm 11}{4}$
• Step 4: Solve for the roots:
  $x = \frac{-7 + 11}{4} = 1$
  $x = \frac{-7 - 11}{4} = -\frac{18}{4} = -4.5$
• Step 5: The roots are $x = 1$ and $x = -4.5$. These roots divide the real number line into intervals: $(-\infty, -4.5)$, $(-4.5, 1)$, and $(1, \infty)$.
• Step 6: Test each interval:
• Interval $(-\infty, -4.5)$: Choose a test point like $x = -5$.
  Calculate $y$ using $x = -5$:
  $y = 2(-5)^2 + 7(-5) - 9 = 50 - 35 - 9 = 6 \, (> 0)$. The function is positive in this interval.
• Interval $(-4.5, 1)$: Choose a test point like $x = 0$.
  Calculate $y$ using $x = 0$:
  $y = 2(0)^2 + 7(0) - 9 = -9 \, (< 0)$. The function is negative in this interval.
• Interval $(1, \infty)$: Choose a test point like $x = 2$.
  Calculate $y$ using $x = 2$:
  $y = 2(2)^2 + 7(2) - 9 = 8 + 14 - 9 = 13 \, (> 0)$. The function is positive in this interval.

The positive domain of $y$ is $(-\infty, -4.5) \cup (1, \infty)$.
The negative domain of $y$ is $(-4.5, 1)$.

Thus, the domains are as follows:
$x < 0 : -4\frac{1}{2} < x < 1$
$x > 1$ or $x > 0 : x < -4\frac{1}{2}$

To find the positive and negative domains of the function $y = -3x^2 + 5x - 2$, we follow these steps:

• Step 1: Find the roots of the equation by solving $-3x^2 + 5x - 2 = 0$.
• Step 2: The quadratic equation $ax^2 + bx + c = 0$ has roots $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Step 3: Calculate the discriminant, $b^2 - 4ac = 5^2 - 4(-3)(-2) = 25 - 24 = 1$.
• Step 4: Find the roots: $x = \frac{-5 \pm \sqrt{1}}{2(-3)} = \frac{-5 \pm 1}{-6}$.
• Step 5: This gives roots $x = \frac{-4}{-6} = \frac{2}{3}$ and $x = \frac{-6}{-6} = 1$.
• Step 6: Determine intervals: $(-\infty, \frac{2}{3})$, $(\frac{2}{3}, 1)$, $(1, \infty)$.
• Step 7: Test a point from each interval in the original equation to determine sign:
• For $x \in (-\infty, \frac{2}{3})$, choose $x = 0$: $y = -3(0)^2 + 5(0) - 2 = -2$ (negative).
• For $x \in (\frac{2}{3}, 1)$, choose $x = 0.8$: $y = -3(0.8)^2 + 5(0.8) - 2 = -1.28 + 4 - 2 = 0.72$ (positive).
• For $x \in (1, \infty)$, choose $x = 2$: $y = -3(2)^2 + 5(2) - 2 = -12 + 10 - 2 = -4$ (negative).

Therefore, the positive domains of the function are when $\frac{2}{3} < x < 1$, and the negative domains are when $x < \frac{2}{3}$ or $x > 1$.

Thus, the solution to the problem is:

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

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

To solve this problem, we'll analyze the given quadratic function $y = \frac{1}{2}x^2 + 2x + 3$ and determine where it is positive and where it is negative.

Step 1: Calculate the discriminant to find out if there are real roots.

The quadratic equation has coefficients $a = \frac{1}{2}$, $b = 2$, $c = 3$. The discriminant $\Delta$ is given by:

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

Since the discriminant is negative ($\Delta < 0$), the quadratic equation does not have real roots; thus, it does not cross the x-axis.

Step 2: Determine the nature of the parabola.

The parabola opens upwards because the leading coefficient $a = \frac{1}{2}$ is positive.

Step 3: Conclude based on the parabola's direction and lack of real roots.

Because the parabola opens upwards and does not intersect the x-axis, the function $y$ is positive for all $x$.

Therefore, the positive domain of the function is for all $x$, and there is no negative domain.

Conclusion:
$x > 0 :$ 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 solve this problem, let's start by finding the roots of the quadratic equation:

The given function is $y = -4x^2 + x + 3$. We set it to zero to find the roots:

$-4x^2 + x + 3 = 0$

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = -4$, $b = 1$, and $c = 3$:

$x = \frac{-1 \pm \sqrt{1^2 - 4(-4)(3)}}{2(-4)}$

$x = \frac{-1 \pm \sqrt{1 + 48}}{-8}$

$x = \frac{-1 \pm \sqrt{49}}{-8}$

$x = \frac{-1 \pm 7}{-8}$

The roots are:

• $x_1 = \frac{-1 + 7}{-8} = \frac{6}{-8} = -\frac{3}{4}$
• $x_2 = \frac{-1 - 7}{-8} = \frac{-8}{-8} = 1$

These roots divide the number line into intervals: $(-\infty, -\frac{3}{4})$, $(-\frac{3}{4}, 1)$, and $(1, \infty)$.

We test each interval to determine where the function is positive or negative:

Interval $(-\infty, -\frac{3}{4})$: Choose $x = -1$.

• $y = -4(-1)^2 + (-1) + 3 = -4 - 1 + 3 = -2$ (Negative)

Interval $(-\frac{3}{4}, 1)$: Choose $x = 0$.

• $y = -4(0)^2 + 0 + 3 = 3$ (Positive)

Interval $(1, \infty)$: Choose $x = 2$.

• $y = -4(2)^2 + 2 + 3 = -16 + 2 + 3 = -11$ (Negative)

Therefore, the function is positive in the interval $(-\frac{3}{4}, 1)$ and negative in the intervals $(-\infty, -\frac{3}{4})$ and $(1, \infty)$.

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

$x > 0 : -\frac{3}{4} < x < 1$

$x > 1$ or $x < 0 : x < -\frac{3}{4}$

The correct answer choice corresponds to:

$x > 1$ or $x<0:x<-\frac{3}{4}$

$x > 0 : -\frac{3}{4} < x <1$