Product Representation: With fractions

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

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 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 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 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 $x$.

To solve this problem, let's examine the function $y = -\left(\frac{1}{3}x + 15\right)^2$ and determine when $y < 0$.

Step 1: Analyze the expression inside the function.
The function involves the square of a linear term, $\left(\frac{1}{3}x + 15\right)$. The square of any real number is always non-negative, meaning $\left(\frac{1}{3}x + 15\right)^2 \geq 0$.

Step 2: Consider the effect of multiplying by $-1$.
When this non-negative square is multiplied by $-1$, the result is always non-positive: $-\left(\frac{1}{3}x + 15\right)^2 \leq 0$.

Step 3: Identify when the function $y$ is less than zero.
For the function to be strictly less than zero, the squared term must be strictly greater than zero: $\left(\frac{1}{3}x + 15\right)^2 > 0$.

Step 4: Determine the zero point to exclude it.
The expression $\left(\frac{1}{3}x + 15\right)^2 = 0$ only when $\frac{1}{3}x + 15 = 0$. Solving this equation gives:

• $\frac{1}{3}x + 15 = 0$
• $\frac{1}{3}x = -15$
• $x = -45$

This means that the function $y = 0$ at $x = -45$. To satisfy $y < 0$, $x$ must be any value other than $-45$.

Therefore, the condition for which the function value is negative is: $x \neq -45$.

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 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 this problem, we'll begin by finding the roots of the quadratic equation $y = \left(x - \frac{1}{2}\right)\left(-x + 3\frac{1}{2}\right)$.

First, set each factor equal to zero:

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

This means the roots of the quadratic are $x = \frac{1}{2}$ and $x = 3\frac{1}{2}$.

Next, analyze the intervals determined by these roots:

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

Perform a sign test within these intervals:

• For $x < \frac{1}{2}$: Both $x - \frac{1}{2}$ and $-x + 3\frac{1}{2}$ are negative, thus their product is positive (negative times negative is positive).
• For $\frac{1}{2} < x < 3\frac{1}{2}$: The factor $x - \frac{1}{2}$ is positive and $-x + 3\frac{1}{2}$ is positive as well, thus product is positive (positive times positive is positive).
• For $x > 3\frac{1}{2}$: $x - \frac{1}{2}$ is positive, but $-x + 3\frac{1}{2}$ is negative, so the product is negative (positive times negative is negative).

Therefore, the quadratic function is negative for: $x > 3\frac{1}{2}$ and $x < \frac{1}{2}$.

The solution to the problem is: $x > 3\frac{1}{2}$ or $x < \frac{1}{2}$.

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'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'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'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'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 $x\ne\frac{1}{5}$.