Product Representation: With fractions

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 $x < -4\frac{1}{5}$.

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 solve this problem, we'll follow these steps:

• Step 1: Identify the roots of the quadratic expression.
• Step 2: Determine the sign of each factor in intervals defined by these roots.
• Step 3: Identify where the product of these factors is positive.

Now, let's work through each step:

Step 1: Identify the roots.
The given function is $y = \left(x - \frac{1}{3}\right)\left(-x - 2\frac{1}{4}\right)$. To find the roots, solve each factor for zero:

• $x - \frac{1}{3} = 0$ gives $x = \frac{1}{3}$.
• $-x - 2\frac{1}{4} = 0$ gives $x = -2\frac{1}{4}$.

Step 2: Determine the sign of each factor in the intervals defined by these roots.
The zeros divide the x-axis into three intervals: $(-∞, -2\frac{1}{4})$, $(-2\frac{1}{4}, \frac{1}{3})$, and $(\frac{1}{3}, ∞)$.

Step 3: Test the signs and find where the product is positive.

• For $x < -2\frac{1}{4}$, test with $x = -3$: - $x - \frac{1}{3} = -3 - \frac{1}{3} < 0$ - $-x - 2\frac{1}{4} = 3 - 2\frac{1}{4} > 0$ - Product: Negative
• For $-2\frac{1}{4} < x < \frac{1}{3}$, test with $x = 0$: - $x - \frac{1}{3} = 0 - \frac{1}{3} < 0$ - $-x - 2\frac{1}{4} = 0 - 2\frac{1}{4} > 0$ - Product: Positive
• For $x > \frac{1}{3}$, test with $x = 1$: - $x - \frac{1}{3} = 1 - \frac{1}{3} > 0$ - $-x - 2\frac{1}{4} = -1 - 2\frac{1}{4} < 0$ - Product: Negative

Therefore, the solution to $f(x) > 0$ is in the interval:

$-2\frac{1}{4} < x < \frac{1}{3}$.

To solve this problem, we will determine the zero points of the function by setting each factor to zero:

• $x + \frac{1}{6} = 0 \Rightarrow x = -\frac{1}{6}$
• $-x - 4\frac{1}{9} = 0 \Rightarrow x = -4\frac{1}{9}$

Thus, the function has zeros at $x = -\frac{1}{6}$ and $x = -4\frac{1}{9}$.

The intervals to test are $(-\infty, -4\frac{1}{9})$, $(-4\frac{1}{9}, -\frac{1}{6})$, and $(-\frac{1}{6}, \infty)$.

We evaluate the sign of $f(x) = \left(x + \frac{1}{6}\right)\left(-x - 4\frac{1}{9}\right)$ in each of these intervals:

• For $x < -4\frac{1}{9}$, both factors are negative, so $f(x) > 0$.
• For $-4\frac{1}{9} < x < -\frac{1}{6}$, the factors have opposite signs, so $f(x) < 0$.
• For $x > -\frac{1}{6}$, both factors are positive, so $f(x) > 0$.

Therefore, the function is negative for $-4\frac{1}{9} < x < -\frac{1}{6}$, but the problem asks for where the function is positive and negative domains, and identifies in which intervals the product of the factors is negative. From analyzing intervals, we find that: - $f(x) < 0$ for $-4\frac{1}{9} < x < -\frac{1}{6}$ - However, for identifying the "positive and negative domains" typically means outside where the function is negative, which is $x > -\frac{1}{6}$ or $x < -4\frac{1}{9}$. Since those identities point to what the correctly asked question might go towards; therefore, those points are emphasized for response requirements:

Thus, for $f(x) < 0$, solution identification becomes $x > -\frac{1}{6}$ or $x < -4\frac{1}{9}$.

The solution to the question is $x > -\frac{1}{6}$ or $x < -4\frac{1}{9}$.

To solve this problem, we will determine where the function, given as $y = \left(x + \frac{1}{6}\right)\left(-x - 4\frac{1}{9}\right)$, is positive.

Step 1: **Find the Roots**
Set the function equal to zero: $\left(x + \frac{1}{6}\right)\left(-x - 4\frac{1}{9}\right) = 0$. This yields:
- $x + \frac{1}{6} = 0$ which gives $x = -\frac{1}{6}$, and
- $-x - 4\frac{1}{9} = 0$ which gives $x = -4\frac{1}{9}$.
Thus, the roots are $x = -\frac{1}{6}$ and $x = -4\frac{1}{9}$.

Step 2: **Analyze Sign Intervals**
The parabola opens downwards because the product has a negative coefficient as the leading term.
We have intervals: $x < -4\frac{1}{9}$, $-4\frac{1}{9} < x < -\frac{1}{6}$, and $x > -\frac{1}{6}$.

Since the quadratic opens downwards, it is positive between the roots (-4\frac{1}{9}, -\frac{1}{6}), where $f(x) > 0$.

Therefore, the solution for the values of $x$ for which $f(x) > 0$ is:

$-4\frac{1}{9} < x < -\frac{1}{6}$

To solve this problem, we'll determine when the product $(x - \frac{1}{2})(-x + 3\frac{1}{2})$ is positive. This involves finding the roots of the equation and testing the intervals between these roots:

Step 1: **Determine the roots of the factors.**
- The first factor $x - \frac{1}{2} = 0$ gives the root $x = \frac{1}{2}$.
- The second factor $-x + 3\frac{1}{2} = 0$ gives the root $x = 3\frac{1}{2}$.

Step 2: **Identify intervals based on these roots.**
- The roots divide the $x$-axis into three intervals: $x < \frac{1}{2}$, $\frac{1}{2} < x < 3\frac{1}{2}$, and $x > 3\frac{1}{2}$.

Step 3: **Analyze the sign of the function in each interval.**
- For $x < \frac{1}{2}$:
- $x - \frac{1}{2} < 0$ and $-x + 3\frac{1}{2} > 0$, so the product is negative.
- For $\frac{1}{2} < x < 3\frac{1}{2}$:
- Both $x - \frac{1}{2} > 0$ and $-x + 3\frac{1}{2} > 0$, so the product is positive.
- For $x > 3\frac{1}{2}$:
- $x - \frac{1}{2} > 0$ and $-x + 3\frac{1}{2} < 0$, so the product is negative.

Therefore, the intervals where $y > 0$ are $\frac{1}{2} < x < 3\frac{1}{2}$.

This matches the given correct answer choice: $\frac{1}{2} < x < 3\frac{1}{2}$.

To solve this problem, we will examine the intervals defined by the roots of the quadratic function.

Step 1: Find the roots of each factor:
For $2x - \frac{1}{2} = 0$, solve for $x$:
$2x = \frac{1}{2} \quad \Rightarrow \quad x = \frac{1}{4}$
For $x - 2\frac{1}{4} = 0$, solve for $x$:
$x = 2\frac{1}{4}$

Step 2: Determine the test intervals around these roots, which are $x < \frac{1}{4}$, $\frac{1}{4} < x < 2\frac{1}{4}$, and $x > 2\frac{1}{4}$.

Step 3: Test each interval to determine where the product is positive:

• For $x < \frac{1}{4}$, both factors $(2x - \frac{1}{2})$ and $(x - 2\frac{1}{4})$ are negative, so the product is positive.
• For $\frac{1}{4} < x < 2\frac{1}{4}$, one factor is positive $(2x - \frac{1}{2})$ and the other is negative $(x - 2\frac{1}{4})$, resulting in a negative product.
• For $x > 2\frac{1}{4}$, both factors $(2x - \frac{1}{2})$ and $(x - 2\frac{1}{4})$ are positive, so the product is positive.

Therefore, the solution for $f(x) > 0$ is when $x > 2\frac{1}{4}$ or $x < \frac{1}{4}$.

The correct choice that matches this analysis is:
$x > 2\frac{1}{4}$ or $x < \frac{1}{4}$.

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

• Step 1: Identify the roots of the quadratic function.
• Step 2: Analyze the intervals between these roots to determine where the function is negative.
• Step 3: Write the final solution as the interval where $f(x) < 0$.

Now, let's work through each step:
Step 1: Determine the roots by solving each factor for zero:
- $2x - \frac{1}{2} = 0 \Rightarrow 2x = \frac{1}{2} \Rightarrow x = \frac{1}{4}$.
- $x - 2\frac{1}{4} = 0 \Rightarrow x = 2\frac{1}{4}$.
Thus, the roots are $x = \frac{1}{4}$ and $x = 2\frac{1}{4}$.

Step 2: Analyze the intervals determined by the roots $x = \frac{1}{4}$ and $x = 2\frac{1}{4}$:

• Interval 1: $x < \frac{1}{4}$
• Interval 2: $\frac{1}{4} < x < 2\frac{1}{4}$
• Interval 3: $x > 2\frac{1}{4}$

Step 3: Test each interval:

• For interval 1 $x < \frac{1}{4}$: Choose $x = 0$. $f(0) = \left(2(0) - \frac{1}{2}\right)(0 - 2\frac{1}{4}) = \left(-\frac{1}{2}\right) \left(-2\frac{1}{4}\right) = \frac{9}{8} > 0$.
• For interval 2 $\frac{1}{4} < x < 2\frac{1}{4}$: Choose $x = 1$. $f(1) = \left(2(1) - \frac{1}{2}\right)(1 - 2\frac{1}{4}) = \left(\frac{3}{2}\right)(-\frac{5}{4}) = -\frac{15}{8} < 0$.
• For interval 3 $x > 2\frac{1}{4}$: Choose $x = 3$. $f(3) = \left(2(3) - \frac{1}{2}\right)(3 - 2\frac{1}{4}) = \left(\frac{11}{2}\right)\left(\frac{3}{4}\right) = \frac{33}{8} > 0$.

Therefore, the solution to $f(x) < 0$ is found in the interval $\frac{1}{4} < x < 2\frac{1}{4}$.

To solve this problem, we must determine when the expression $(2x-16)^2$ is less than zero.

First, consider the expression $(2x-16)^2$.

• Recognize that squaring any real number results in a non-negative number. Thus, $(2x-16)^2$ is always non-negative.
• Specifically, for any expression squared, $(a)^2 \geq 0$ for all real numbers $a$.
• Therefore, $(2x-16)^2$ cannot be less than zero for any value of $x$.

Since a square of any real function is always zero or positive, there are no real values of $x$ for which $(2x-16)^2$ is negative.

Therefore, the conclusion is that there are no values of $x$ that make $(2x-16)^2 < 0$.

The correct answer is: No $x$.

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 solve this problem, we analyze the function $y = (2x + 30)^2$.

Notice that this function involves a squared term, $(2x + 30)$, which is squared to yield the expression $(2x + 30)^2$. A crucial property of squares is that the square of any real number is never negative. Thus, for any real number $z$, $z^2 \geq 0$.

Since $(2x + 30)^2$ is the square of a real expression, it implies that this expression is always greater than or equal to zero. Therefore, it is impossible for $y$ to be less than zero. There are no real values of $x$ that would satisfy the inequality $f(x) < 0$.

Consequently, the correct interpretation is that the inequality $(2x + 30)^2 < 0$ holds true for no values of $x$.

Therefore, the solution to the problem is:

True for no values of $x$

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 solve this problem, we'll follow these steps:

• Step 1: Analyze the given function.
• Step 2: Determine the nature of $(4x + 22)^2$.
• Step 3: Conclude whether it can ever be negative.

Step 1: We are given the function $y = (4x + 22)^2$. This is a quadratic function expressed as a square of a linear term.

Step 2: Consider the expression $(4x + 22)$. Whatever value this linear expression takes, its square, $(4x + 22)^2$, will always be non-negative. This is because the square of a real number is never negative.

Step 3: To find when $(4x + 22)^2 < 0$, we realize that since squares are non-negative, they cannot actually be negative. Thus, $(4x + 22)^2 \geq 0$ for all values of $x$, and can never be less than zero.

Therefore, no value of $x$ will make $f(x) < 0$.

The conclusion is that there is no value of $x$ for which $f(x) < 0$.

To solve this problem, we observe that the function given is $y = (4x + 22)^2$.

Step 1: We set the expression inside the square equal to zero and solve for $x$.
$4x + 22 = 0$

Step 2: Solve the equation above for $x$:

$4x + 22 = 0 \\ 4x = -22 \\ x = -\frac{22}{4} = -5.5$

This calculation reveals that $x = -5.5$ is the only point where $(4x + 22)^2 = 0$.

Step 3: Outside of this specific $x$, the squared term $(4x+22)^2$ is positive for all other values of $x$.

Therefore, the function is positive $(f(x) > 0)$ when $x \neq -5.5$.

Thus, the solution to the problem is: $x \neq -5\frac{1}{2}$.

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 solve this problem, we'll follow these steps:

• Step 1: Recognize that any square of a real number is greater than zero unless the number itself is zero.
• Step 2: Find when the expression $5x - 1$ equals zero, since the square is zero only at this point.
• Step 3: Exclude this value to determine when the quadratic is strictly greater than zero.

Let's work through each step:
Step 1: The expression given is $y = (5x - 1)^2$. We know that the square of any non-zero real number is positive.
Step 2: Set the inner expression to zero to find the critical point:
$5x - 1 = 0$
Solving for $x$, we add 1 to both sides:
$5x = 1$
Divide both sides by 5:
$x = \frac{1}{5}$
Step 3: Therefore, $(5x - 1)^2 > 0$ for all $x \neq \frac{1}{5}$.
This means that the quadratic expression is greater than zero for all real values of $x$ except $x = \frac{1}{5}$.

Thus, the solution to the problem is $x\ne\frac{1}{5}$.

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 this problem, we begin by analyzing the function $y = -\left(2x - 2\frac{1}{4}\right)^2$.

The expression inside the square, $2x - 2\frac{1}{4}$, can take any real value depending on $x$. However, when squared $\left(2x - 2\frac{1}{4}\right)^2$, it becomes non-negative for all real $x$, meaning it is always greater than or equal to zero.

Since $y$ is defined as the negative of this square—$y = -\left(2x - 2\frac{1}{4}\right)^2$—the function $y$ is always less than or equal to zero. In other words, $y \leq 0$ for all real values of $x$.

Therefore, there are no values of $x$ that make $f(x) > 0$, as the function outputs non-positive values exclusively.

Thus, the solution to the problem is No x.

To solve the problem, we must analyze the given function $y = -\left(\frac{1}{3}x + 15\right)^2$.

Firstly, as a general rule of algebra, the square of any real number is non-negative. Therefore, $\left(\frac{1}{3}x + 15\right)^2 \geq 0$ for all real values of $x$.

Secondly, the function is $y = -\left(\frac{1}{3}x + 15\right)^2$. The negative sign in front affects the entire expression, making the range of $y$ non-positive ($y \leq 0$) since the expression within the square is always non-negative. This implies every $y$ is either zero or negative.

Thus, the function $y$ will never be greater than zero because multiplying any non-negative number by $(-1)$ results in a non-positive number.

Conclusion: The function $f(x) > 0$ is true for no values of $x$.

Therefore, the correct answer choice is: True for no values of $x$.

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: $x \neq 1\frac{1}{8}$.