Product Representation: Identify the positive and negative domain

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

Solution: We begin by finding the roots of the function $y = (x-4)(x+2)$.

Step 1: Find the roots by solving $(x-4)(x+2) = 0$.

• Root 1: $x - 4 = 0$ implies $x = 4$.
• Root 2: $x + 2 = 0$ implies $x = -2$.

Step 2: The function changes sign at the roots, so we analyze the intervals determined by these roots: $(-\infty, -2)$, $(-2, 4)$, and $(4, \infty)$.

Step 3: Determine where $y > 0$ within these intervals.

• Select a test point from the interval $(-\infty, -2)$, e.g., $x = -3$: $y = (-3-4)(-3+2) = (-7)(-1) = 7$ which is positive.
• Select a test point from the interval $(-2, 4)$, e.g., $x = 0$: $y = (0-4)(0+2) = (-4)(2) = -8$ which is negative.
• Select a test point from the interval $(4, \infty)$, e.g., $x = 5$: $y = (5-4)(5+2) = (1)(7) = 7$ which is positive.

Step 4: Conclude that the function is positive in the intervals $(-\infty, -2)$ and $(4, \infty)$.

Therefore, the solution to the problem is that $f(x) > 0$ when $x < -2$ or $x > 4$.

Upon reviewing the problem's given correct answer, identify any typographical error in it.

Consequently, the function is positive for $x > 4$ or $x < -2$.

The function is given in vertex form: $y = -\left(2x - 2\frac{1}{4}\right)^2$, which translates to $y = -\left(2x - 2.25\right)^2$. The vertex occurs when the expression inside the square is zero, which is at $x = 1\frac{1}{8}$. This is the maximum point due to the negative coefficient, making the function value at the vertex equal to zero.

For $f(x) < 0$, the square term $\left(2x - 2.25\right)^2$ must be non-zero. Thus, set $2x - 2\frac{1}{4} = 0$ to find the $x$ that needs to be excluded:

$2x - 2.25 = 0$

$2x = 2.25$

$x = 1.125$ or $x = 1\frac{1}{8}$

Therefore, for $f(x) < 0$, $x$ should not be equal to $1\frac{1}{8}$.

The correct condition is: $x \neq 1\frac{1}{8}$.

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 $y = (x-6)(x+6)$ can be rewritten as $y = x^2 - 36$. This is a quadratic function, and we need to find where it is positive: $x^2 - 36 > 0$.

First, identify the roots of the quadratic equation $x^2 - 36 = 0$. Solving for $x$, we get:

• $x^2 = 36$
• $x = \pm 6$

Thus, the roots are $x = 6$ and $x = -6$.

Next, examine the intervals determined by these roots: $(-\infty, -6)$, $(-6, 6)$, $(6, \infty)$.

For each interval, we check the sign of $x^2 - 36$ to determine where the expression is positive.

1. **Interval $(-\infty, -6)$:** Choose $x = -7$:
$(x-6)(x+6) = (-7-6)(-7+6) = (-13)(-1) = 13$, which is positive.

2. **Interval $(-6, 6)$:** Choose $x = 0$:
$(x-6)(x+6) = (0-6)(0+6) = (-6)(6) = -36$, which is negative.

3. **Interval $(6, \infty)$:** Choose $x = 7$:
$(x-6)(x+6) = (7-6)(7+6) = (1)(13) = 13$, which is positive.

Therefore, the quadratic $x^2 - 36 > 0$ in the intervals $(-\infty, -6)$ and $(6, \infty)$. The function is positive on these intervals.

Since the solution matches choice id="4", the values of $x$ for which $f(x) > 0$ are:
$x > 6$ or $x < -6$.

Thus, the solution to the problem is $x > 6$ or $x < -6$.

The given function is $f(x) = (x+1)(x+5)$. We need to determine for which values of $x$ this function is greater than zero.

First, let's find the roots of the function. Set each factor to zero to find the roots:
$x+1 = 0 \rightarrow x = -1$
$x+5 = 0 \rightarrow x = -5$

These roots divide the number line into three intervals: $x < -5$, $-5 < x < -1$, and $x > -1$.

Next, we will determine the sign of $f(x)$ in each interval:

• For $x < -5$, choose a test value like $x = -6$. Both $x+1$ and $x+5$ are negative, so their product $(x+1)(x+5) > 0$.
• For $-5 < x < -1$, choose a test value like $x = -3$. Here $x+1$ is negative, and $x+5$ is positive, making their product $(x+1)(x+5) < 0$.
• For $x > -1$, choose a test value like $x = 0$. Both $x+1$ and $x+5$ are positive, so their product $(x+1)(x+5) > 0$.

Therefore, $f(x) > 0$ when $x < -5$ or $x > -1$.

Thus, the solution is that $f(x) > 0$ for $x > -1$ or $x < -5$.

To determine for which values of $x$ the expression $y = (3x+1)(1-3x)$ is less than zero, we follow these steps:

• Find the roots of each factor:
• The factor $3x+1 = 0$ gives the root $x = -\frac{1}{3}$.
• The factor $1-3x = 0$ gives the root $x = \frac{1}{3}$.
• Determine the intervals created by these roots:
• Interval 1: $x < -\frac{1}{3}$
• Interval 2: $-\frac{1}{3} < x < \frac{1}{3}$
• Interval 3: $x > \frac{1}{3}$
• Test the sign of the product within each interval:
• For $x < -\frac{1}{3}$, choose $x = -1$:
• $3x+1$ is negative, and $1-3x$ is positive. Product = negative.
• For $-\frac{1}{3} < x < \frac{1}{3}$, choose $x = 0$:
• $3x+1$ is positive, and $1-3x$ is positive. Product = positive.
• For $x > \frac{1}{3}$, choose $x = 1$:
• $3x+1$ is positive, and $1-3x$ is negative. Product = negative.

Conclusion: The product $(3x+1)(1-3x)$ is less than zero for:

x > \frac{1}{3} or x < -\frac{1}{3}

To determine for which values of $x$ the function $y = -\left(4x + 32\right)^2$ is positive, we must analyze the behavior of this function.

The function $y = -\left(4x + 32\right)^2$ is a quadratic function with a negative leading coefficient (the negative sign outside the squared term). This indicates that the parabola opens downwards. Let's break down the expression:

• The expression inside the parentheses is $4x + 32$.
• Squaring any real number $\left( 4x + 32 \right)^2$ always results in a non-negative value (it is zero or positive).
• Multiplying this non-negative result by $-1$ makes $y$ non-positive (either zero or negative).

Since the smallest value $\left( 4x + 32 \right)^2$ can be is zero (when $x = -8$), and all other values will just make it negative when multiplied by $-1$, $y$ can never be greater than zero for any real number $x$.

Thus, there are no values of $x$ for which $f(x) > 0$. Consequently, the answer is:

True for no values of $x$.

To determine for which values of $x$ the function $y = (x + 6)(x - 3)$ is positive, let's work through the steps:

• Step 1: Identify the roots of the quadratic equation. Set the equation equal to zero: $(x + 6)(x - 3) = 0$.
  The roots are $x = -6$ and $x = 3$.
• Step 2: Analyze the sign of the function in the intervals around the roots:
  - Interval 1: $x < -6$
  - Interval 2: $-6 < x < 3$
  - Interval 3: $x > 3$
• Step 3: Test the sign of the quadratic expression in each interval:
• In Interval 1 ($x < -6$): Choose $x = -7$:
  $(x + 6) = -7 + 6 = -1$; $(x - 3) = -7 - 3 = -10$
  The product $(x + 6)(x - 3) = (-1)(-10) = 10$, which is positive.
• In Interval 2 ($-6 < x < 3$): Choose $x = 0$:
  $(x + 6) = 0 + 6 = 6$; $(x - 3) = 0 - 3 = -3$
  The product $(x + 6)(x - 3) = (6)(-3) = -18$, which is negative.
• In Interval 3 ($x > 3$): Choose $x = 4$:
  $(x + 6) = 4 + 6 = 10$; $(x - 3) = 4 - 3 = 1$
  The product $(x + 6)(x - 3) = (10)(1) = 10$, which is positive.

Step 4: Conclusion:
The inequality $f(x) > 0$ holds for $x$ in Interval 1 ($x < -6$) and Interval 3 ($x > 3$).
Therefore, the values of $x$ that satisfy $f(x) > 0$ are $x < -6$ or $x > 3$.

The solution to the problem is $x > 3$ or $x < -6$.

Thus, the correct choice among the provided options is Choice 4.

To determine the values of $x$ for which the function $y = (3x + 30)^2$ is greater than zero, consider the following steps:

• Step 1: Recognize the structure of the function. The function is of the form $(\text{expression})^2$. For the function to be greater than zero, the expression inside the square must not equal zero.
• Step 2: Solve $3x + 30 = 0$ to find when the function equals zero.
  Subtract 30 from both sides: $3x = -30$.
  Divide by 3: $x = -10$.
• Step 3: Exclude $x = -10$ from the domain where the function is greater than 0. For all $x \neq -10$, $(3x + 30)^2$ is positive because it results from squaring a non-zero real number.

Therefore, $f(x) > 0$ for all $x \neq -10$.

The correct answer is $x\ne-10$.

To determine when the function $y = (2x - 16)^2$ is greater than zero, we observe the following:

• The function's expression, $(2x - 16)^2$, is a square and thus always non-negative ($\geq 0$).
• The expression will equate to zero when the inside term is zero: $2x - 16 = 0$.
• Solve the equation $2x - 16 = 0$ to find $x = 8$.
• Therefore, $(2x - 16)^2 = 0$ only at $x = 8$.
• For $y > 0$, $(2x - 16)^2$ must be greater than zero, which occurs for all $x$ except $x = 8$.

The solution is $x \ne 8$, which means $y$ is positive for all $x$ except $x = 8$.

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 $2.3 < x < 4.4$.

To find the intervals where $f(x) = (x + 1)(6 - x) > 0$, follow these steps:

• Step 1: Find the roots by setting each factor to zero.
  The roots occur at $x + 1 = 0$ and $6 - x = 0$. Solving these gives $x = -1$ and $x = 6$.
• Step 2: Determine the sign of the function in the intervals defined by these roots: $(-\infty, -1)$, $(-1, 6)$, and $(6, \infty)$.
• Step 3: Evaluate the sign of the product in each interval:
  - For $x \in (-\infty, -1)$, choose $x = -2$. Then, both factors $(x + 1)$ and $(6 - x)$ are negative, making their product positive.
  - For $x \in (-1, 6)$, choose $x = 0$. Then $(x + 1)$ is positive and $(6 - x)$ is positive, making their product positive.
  - For $x \in (6, \infty)$, choose $x = 7$. Then $(x + 1)$ is positive, but $(6 - x)$ is negative, making their product negative.

Thus, the intervals where $f(x) > 0$ are $x < -1$ and $x > 6$.

The solution to the problem is $x > 6$ or $x < -1$.

To find the set of $x$ values where $y = \left(\frac{1}{3}x-\frac{1}{6}\right)\left(-x-4\frac{1}{5}\right)$ is positive, we need to determine where each factor changes sign.

First, find the zeros of the linear factors:

• For $\frac{1}{3}x - \frac{1}{6} = 0$:
  Solving gives $\frac{1}{3}x = \frac{1}{6}$ or $x = \frac{1}{2}$.
• For $-x - 4\frac{1}{5} = 0$:
  Solving gives $-x = -4\frac{1}{5}$ or $x = -4\frac{1}{5}$.

These zeros split the real number line into three intervals. Let's determine the sign of each expression in the intervals:

• Interval $(-\infty, -4\frac{1}{5})$:
  - $\frac{1}{3}x-\frac{1}{6} < 0$ and $-x-4\frac{1}{5} > 0$
• Interval $(-4\frac{1}{5}, \frac{1}{2})$:
  - $\frac{1}{3}x-\frac{1}{6} < 0$ and $-x-4\frac{1}{5} < 0$
• Interval $(\frac{1}{2}, \infty)$:
  - $\frac{1}{3}x-\frac{1}{6} > 0$ and $-x-4\frac{1}{5} < 0$

The product $\left(\frac{1}{3}x-\frac{1}{6}\right)\left(-x-4\frac{1}{5}\right)$ is positive in the interval where both factors are negative or both are positive:

Therefore, the solution is $-4\frac{1}{5} < x < -\frac{1}{2}$, matching with choice 3.

To find when the function $y = \left(x - \frac{1}{2}\right)\left(x + 6\frac{1}{2}\right)$ is positive, we proceed as follows:

First, identify the roots of the expression by solving $x - \frac{1}{2} = 0$ and $x + 6\frac{1}{2} = 0$. These calculations give us the roots $x = \frac{1}{2}$ and $x = -6\frac{1}{2}$, or $x = -\frac{13}{2}$.

Next, determine the sign of the product $(x - \frac{1}{2})(x + \frac{13}{2})$ over the intervals defined by these roots:

• Interval 1: $x < -\frac{13}{2}$. In this region, both $x - \frac{1}{2}$ and $x + \frac{13}{2}$ are negative, so their product is positive.
• Interval 2: $-\frac{13}{2} < x < \frac{1}{2}$. In this region, $x + \frac{13}{2}$ is positive and $x - \frac{1}{2}$ is negative, so their product is negative.
• Interval 3: $x > \frac{1}{2}$. In this region, both $x - \frac{1}{2}$ and $x + \frac{13}{2}$ are positive, so their product is positive.

Therefore, the function is positive for $x < -6\frac{1}{2}$ and $x > \frac{1}{2}$.

Thus, the solution is:
$x > \frac{1}{2}$ or $x < -6\frac{1}{2}$

To find where $f(x) = (5x - 1)^2 < 0$, we start by recognizing a fundamental property of squares:

• The square of any real number is always non-negative. Therefore, $(5x - 1)^2 \geq 0$ for all real $x$.

This implies that $(5x - 1)^2$ can never be less than zero for any real value of $x$.

The inequality $(5x - 1)^2 < 0$ has no solution in the real numbers.

Therefore, there are no values of $x$ for which $f(x) < 0$ is true.

So the logical conclusion is: True for no values of $x$.

To identify the range of $x$ such that $y = (3x + 3)(2 - x) < 0$, we'll follow these steps:

• Step 1: Find the roots of the equation by solving $(3x + 3) = 0$ and $(2 - x) = 0$.
• Step 2: Find the intervals created by these roots.
• Step 3: Test each interval to determine the sign of the product.

Let's execute each step:

Step 1: Solving the equations:
First root: Set $3x + 3 = 0$ which gives $x = -1$.
Second root: Set $2 - x = 0$ which gives $x = 2$.

Step 2: The roots divide the real number line into three intervals:

• $x < -1$
• $-1 < x < 2$
• $x > 2$

Step 3: Analyze each interval:

- For $x < -1$: Choose $x = -2$. The expression becomes $(3(-2) + 3)(2 - (-2)) = (-3)(4) = -12$, which is negative.

- For $-1 < x < 2$: Choose $x = 0$. The expression becomes $(3(0) + 3)(2 - 0) = (3)(2) = 6$, which is positive.

- For $x > 2$: Choose $x = 3$. The expression becomes $(3(3) + 3)(2 - 3) = (12)(-1) = -12$, which is negative.

Therefore, the function is negative for $x < -1$ or $x > 2$.

The solution to this problem is $x > 2$ or $x < -1$.

To solve the problem of determining for which values of $x$ the function $y = (5x-1)(4x-\frac{1}{4})$ is negative, we will follow these steps:

• Step 1: Determine the roots of the function by setting each factor equal to zero.
• Step 2: Analyze the sign of the function on intervals defined by these roots.
• Step 3: Identify where the function is negative based on this analysis.

Let's proceed with these steps:

Step 1: Find the roots of the function.

To find the roots, set each factor equal to zero:

$5x - 1 = 0 \Rightarrow x = \frac{1}{5}$

$4x - \frac{1}{4} = 0 \Rightarrow x = \frac{1}{16}$

Step 2: Determine the intervals on the number line.

The roots divide the number line into the following intervals: $(-\infty, \frac{1}{16})$, $(\frac{1}{16}, \frac{1}{5})$, and $(\frac{1}{5}, \infty)$.

Step 3: Analyze the sign of the function in each interval:

• For $x < \frac{1}{16}$: Both $5x - 1$ and $4x - \frac{1}{4}$ are negative. Hence, their product is positive.
• For $\frac{1}{16} < x < \frac{1}{5}$: Here, $5x - 1$ is negative, and $4x - \frac{1}{4}$ is positive, making the product negative.
• For $x > \frac{1}{5}$: Both $5x - 1$ and $4x - \frac{1}{4}$ are positive, resulting in a positive product.

Now, consolidate the findings:

The function $y = (5x-1)(4x-\frac{1}{4})$ is less than zero for values $\frac{1}{16} < x < \frac{1}{5}$.

Therefore, the solution to the given problem is $\frac{1}{16} < x < \frac{1}{5}$.

To solve the problem, we analyze the function:

The function is given as $y = (3x + 1)(1 - 3x)$. This function is a quadratic expression in the factored form, which allows us to find the roots and analyze the intervals.

Step 1: Identify the roots.
Set each factor equal to zero:
$3x + 1 = 0$ leads to $x = -\frac{1}{3}$.
$1 - 3x = 0$ leads to $x = \frac{1}{3}$.

Step 2: Determine the sign in each interval divided by the roots.
The roots divide the real number line into the following intervals: $(-\infty, -\frac{1}{3})$, $(-\frac{1}{3}, \frac{1}{3})$, and $(\frac{1}{3}, \infty)$.

Step 3: Test the sign of $y$ in each interval:

• For $x \in (-\infty, -\frac{1}{3})$, choose $x = -1$:
  $(3(-1) + 1)(1 - 3(-1)) = (-2)(4) = -8$. So, $y < 0$.
• For $x \in (-\frac{1}{3}, \frac{1}{3})$, choose $x = 0$:
  $(3(0) + 1)(1 - 3(0)) = (1)(1) = 1$. So, $y > 0$.
• For $x \in (\frac{1}{3}, \infty)$, choose $x = 1$:
  $(3(1) + 1)(1 - 3(1)) = (4)(-2) = -8$. So, $y < 0$.

Thus, the function $y = (3x + 1)(1 - 3x)$ is positive for $x$ in the interval $-\frac{1}{3} < x < \frac{1}{3}$.

Therefore, the values of $x$ for which $f(x) > 0$ are $-\frac{1}{3} < x < \frac{1}{3}$.

The correct answer is: $-\frac{1}{3} < x < \frac{1}{3}$.