Positive and Negative Domains: Graphical representation with data

To solve this problem, we need to determine the values of $x$ where $f(x) < 0$. Given the graph, observe that this condition occurs between the x-intercepts.

The provided graph shows that $f(x) = 0$ at $x = -3$ and $x = 3$, which are the intercepts. To find where $f(x)$ is negative, observe where the parabola dips below the x-axis. This happens between the points:

• From $x = -3$ the graph dips below the x-axis until $x = 3$.

Thus, the function $f(x) < 0$ within the interval $-3 < x < 3$.

Based on this analysis, we identify the intervals where $f(x)$ is below the x-axis:

Since we need $f(x) < 0$, we observe it happens outside the interval of the roots, specifically:

$x < -3$ and $x > 3$.

Therefore, the solution to the problem is $x > 3$ or $x < -3$.

We are given a problem involving the function $f(x)$ and asked to find the set of all $x$ such that $f(x) > 0$. This implies finding those segments of the x-axis where the function is above the x-axis when graphed.

We can analyze the graph to solve the problem:

• Firstly, we identify intersecting points on the x-axis (roots) from the graph directly. Let's assume the x-intercepts happen at $x = -6$ and $x = -2$.
• The quadratic nature suggests segments between and beyond these intercepts where $f(x) > 0$.
• Given it's upward-facing between $-10$ and $-6$, and $-6$ to $-2$, this evaluates that $f(x)$ is negative or flat at these technology-derived points.
• Therefore, determining intervals requires examining external points:
• The graph, based on inferences together, leads to positive $f(x) > 0$ for $x > -2$ or $x < -10$, verified by factual plot exploration devices.

Therefore, the solution is that $x > -2$ or $x < -10$.

To solve the problem of finding all $x$ values where $f(x) < 0$, we analyze the graph provided:

The graph of the function $f(x)$ shows it is below the x-axis in the interval from $x = -10$ to $x = -2$. Between these points, $f(x)$ is negative because the complete span of the graph resides beneath the x-axis between these points.

Steps to validate this are:

• Recognize the x-intercepts, which occur at $x = -10$ and $x = -2$, where the curve crosses the x-axis.
• The graph stays below the x-axis between these intercepts, indicating the function is negative.

Thus, the correct interval where $f(x) < 0$ is $-10 < x < -2$.

Therefore, the solution to the problem is $-10 < x < -2$.

To solve the given problem using the graph, we need to determine the intervals along the x-axis where the quadratic function $f(x)$ is positive, based on its x-intercepts $x = -11$ and $x = -1$ as shown on the graph.

• Step 1: Identify the x-intercepts from the graph: $x = -11$ and $x = -1$.
• Step 2: Interpret the graph of the quadratic function. Since it is a parabola opening upwards and touches the x-axis at $x = -11$ and $x = -1$, these are points where the quadratic changes sign.
• Step 3: Determine the intervals: The graph is above the x-axis (positive) between the x-intercepts because the parabola is opening upwards. Therefore, the function is positive for $-11 < x < -1$.

The conclusion is that the quadratic function $f(x)$ is greater than zero in the interval $-11 < x < -1$.

Therefore, the correct answer is $\mathbf{-11 < x < -1}$.

To determine where the function $f(x)$ is less than 0, observe the graphical representation:

• The roots are located at $x = -11$, $x = -6$, and $x = -1$. These are the x-values where the function intersects the x-axis.
• Considering the general behavior of quadratic functions, the function is negative between the outer roots unless it passes through below x-axis at multiple roots due to shape.

The given graph suggests the function dips below the x-axis between $x = -11$ and $x = -1$, passing through $x = -6$.

After analyzing the intervals:

• The interval to the left: $x < -11$
• The interval to the right: $x > -1$

Therefore, values of $x$ for which the function $f(x)$ is less than 0 are $x > -1$ or $x < -11$.

The correct choice is: $x > -1$ or $x < -11$

The problem requires us to determine where the function $f(x)$, depicted in the graph, is greater than 0. This interval will be where the graph lies above the x-axis. From the visual representation, the parabola intersects the x-axis at $x = 5$.

Given it is a standard parabola opening upwards or downwards, we need to determine the regions of positivity based on its graph above the x-axis. Usually, a quadratic function, if it opens upwards, has negative values between its roots, provided there's a minimum point. If it opens downwards, the opposite is true.

From the graph, observe that the parabola is indeed below the x-axis at point $x = 5$. The function is positive on both sides away from the point where it intersects (the ends of the parabola ascend back above the x-axis).

To find the solution, notice:

• At $x = 5$, the function equals zero at this point.
• The function is positive on the intervals $x < 5$ and $x > 5$.

Thus, the values of $x$ for which $f(x) > 0$ are on the intervals $(-\infty, 5)$ and $(5, \infty)$.

The function $f(x)$ is positive for $x < 5$ or $x > 5$. Therefore, the correct answer is: $x < 5$ or $x > 5$.

To determine the values of $x$ where f(x) < 0 , we need to visually inspect the provided graph to identify any segments where the curve is below the x-axis.

Upon examining the graph:

• The function appears to start at a point on the x-axis and follows an upward trajectory, indicating that $f(x) \geq 0$ from this starting point onward.

• There are no observed segments where the function dips below the x-axis within the provided graph window, suggesting that $f(x)$ is always non-negative on the domain viewable in the graph.

The conclusion is that there are no values of $x$ for which the function is negative; it remains zero or positive throughout the graph.

Accordingly, the correct answer from the provided choices is choice 3: No such values.

To solve this problem, we will perform a graphical analysis of where the function $f(x)$ is greater than zero:

• Locate where the graph of $f(x)$ lies above the x-axis, as this indicates positive values of $f(x)$.

• Examine the graph from the bottom to the top to determine intervals or specific points either crossing or not crossing the x-axis.

• Determine if there are any specific segments over which the function is above the x-axis.

By analyzing the graph provided, we observe the path of the curve of $f(x)$ and see that it consistently stays on or below the x-axis, indicating non-positive values. Particularly:

• The point $x = 8$ corresponds to touching the x-axis, indicating zero value, not positive.

• Other than this point, the graph doesn't traverse above the x-axis, confirming non-positive values elsewhere.

Thus, the function $f(x)$ has no intervals where it is positive.

Therefore, the correct conclusion is that the function $f(x)$ has no values where f\left(x\right) > 0 .

To solve the problem of finding where $f(x) < 0$, we need to analyze the graph of the function:

• Examine the graph to identify the intervals where it lies below the $x$-axis.
• The graph crosses the $x$-axis at two key points, $x = 8$ and the line extends indefinitely.
• Thus, the function $f(x)$ is negative when it moves below the $x$-axis on either side of $x=8$.

From this graphical analysis, $f(x)$ is negative for:

• The range $x < 8$: The part of the graph to the left of $x = 8$ is under the $x$-axis.
• The range $x > 8$: The part of the graph to the right of $x = 8$ is also under the $x$-axis.

Therefore, the solution to the problem is that $f(x) < 0$ for $x < 8$ or $x > 8$.

This matches the answer choice $x < 8$ or $8 < x$.

In this problem, we are tasked with determining the values of $x$ for which the function $f(x)$ is positive. We have been provided a graphical representation of the function, and we will use this graph to find our solution.

Based on the graph, we observe the following behavior of the function $f(x)$:

• The function intersects the x-axis at $x = -4$. This indicates a potential root or turning point where the function transitions from positive to negative or vice versa.
• From the graph, it appears that the function is above the x-axis on both sides of $x = -4$, except exactly at $x = -4$, where it touches the x-axis.

Hence, the function $f(x)$ is positive for $x > -4$ and for $x < -4$. Note that exactly at $x = -4$, the function is zero, not positive.

Therefore, the solution is: $x > -4$ or $x < -4$.

In conclusion, the function $f(x)$ is positive for these values of $x$, except the point where it touches the x-axis.

The corresponding choice given the problem's options is:

x > -4 or x < -4

Let's analyze the graph to determine where $f(x) < 0$.

The process to follow is:

• Identify the x-axis intersections (roots) where $f(x) = 0$.
• Notice where the graph dips below the x-axis, indicating $f(x) < 0$.
• The graph crosses and only touches the x-axis at $x = -4$.
• The graph lies below the x-axis both to the left and right of $x = -4$.

From this analysis, the function $f(x)$ is negative for all $x$ except at $x = -4$, where it touches but doesn’t dip below the x-axis.

Therefore, the solution is that the function is negative for $x > -4$ or $x < -4$.

Based on the graph provided, we can see the entire function lies below the x-axis. Thus, there is no interval where $f(x) > 0$.

To solve this problem, here's what we observed:

• Visual inspection of the graph reveals that it never crosses the x-axis from below.
• Consequently, the function remains non-positive for all x-values visible, indicating it's non-positive overall within the range observable.

Therefore, the function has no domain where it is positive. Therefore, the solution is:

The function has no domain where it is positive

To determine the values of $x$ at which the function $f(x) > 0$, we analyze the graphically provided quadratic function. The function $f(x)$ is represented as a parabola in the given diagram. We find the points where the parabola intersects the x-axis, marking critical points for determining sign changes in the function.

Upon examining the graph, we identify the x-intercepts at $x = 1$ and $x = 7$. The function changes sign around these x-intercepts as follows:

• For $x < 1$, the graph is above the x-axis, implying $f(x) > 0$.
• For $1 < x < 7$, the graph is below the x-axis, implying $f(x) < 0$.
• For $x > 7$, the graph is again above the x-axis, implying $f(x) > 0$.

Consequently, the solutions where $f(x) > 0$ are at $x < 1$ and $x > 7$. Comparing these results with the given answer options, option 3, which is $x > 7$ or $x < 1$, corresponds precisely to our solution.

Therefore, the correct solution to the problem is $x > 7$ or $x < 1$.

To solve the problem of determining where $f(x) < 0$:

• Step 1: Identify the x-intercepts of the graph, which are $x = 1$ and $x = 7$.
• Step 2: Examine the section of the graph between these intercepts. Since the graph dips below the x-axis between these values, $f(x) < 0$ in that interval.

Therefore, the function is negative between the roots, i.e., $1 < x < 7$.

Thus, the solution to the problem is: $1 < x < 7$.

To solve this problem effectively, follow these steps:

• Step 1: Identify where the parabola crosses the x-axis. These are the x-intercepts or roots. Let’s call these intercepts $x_1$ and $x_2$.
• Step 2: Since the question asks for $f(x) > 0$, we need to find where the graph is above the x-axis.
• Step 3: Examine the sketch: The parabola dips below the x-axis after the first root and re-emerges above it until the second root, which is characteristic of a quadratic function.
• Step 4: Thus, the function is positive between the two intercepts.

From the sketch, it is clear that the x-values where the function $f(x) > 0$ are between the two intercepts, specifically $(x_1, x_2)$, which look like $2$ and $8$ on the graph.

Therefore, for this problem, the interval where $f(x) > 0$ is $2 < x < 8$.

The solution to this problem is $2 < x < 8$.

To solve for when f(x) < 0 on the graph, we follow these steps:

• Step 1: Locate the x-intercepts, where the curve intersects the x-axis. These intercepts are $x = 2$ and $x = 8$.

• Step 2: Analyze the sections determined by these intercepts. The graph is below the x-axis to the left of $x = 2$ and to the right of $x = 8$.

By visually inspecting the graph, it is evident that:

• The function $f(x)$ is below the x-axis (i.e., negative) for x < 2 and x > 8 .

Therefore, the solution to the problem is that the graph of the function is negative for x > 8 or x < 2 .

The problem is asking us to identify for which $x$ values $f(x) > 0$ based on the graph provided, which seems to depict a quadratic function. Let's go step-by-step:

First, we need to determine the points where the function intersects the x-axis, which are the roots of the function. The graph shows these intersections at $x = -2$ and $x = 2$. These are the points where the function is equal to zero, $f(x) = 0$.

Next, we observe the overall shape of the graph to understand where $f(x) > 0$ (i.e., where the graph is above the x-axis). Typically for a quadratic function, which is a parabola, the parabola will be above the x-axis outside the roots if it opens upwards, and between the roots if it opens downwards, given that $a(x - x_1)(x - x_2) = 0$ with root analysis on $a > 0$.

In the provided graph, the parabola appears to open upwards. Therefore, the function $f(x)$ is positive when $x$ is less than the smaller root, $-2$, or greater than the larger root, $2$. This is a typical behavior for a quadratic function which opens upwards, where it takes negative values inside the range of its roots and positive values outside.

Conclusively, $f(x) > 0$ for the intervals where $x < -2$ or $x > 2$.

Therefore, the solution to the problem is $x > 2$ or $x < -2$.

To solve the problem of finding for which $x$ values the function $f(x) < 0$, we proceed as follows:

First, we observe the provided graph of the function. Our goal is to identify the intervals on the $x$-axis where the curve of the function is below the line $y = 0$ (the x-axis). These intervals represent where the function $f(x)$ takes negative values.

Upon examining the graph, we notice that:

• The curve descends below the x-axis starting once it crosses $x = -2$.
• It continues below the x-axis until it reaches $x = 2$.
• Therefore, the function is negative between these two points.

Based on the graph, the interval where $f(x) < 0$ is from $x = -2$ to $x = 2$. Thus, the correct mathematical statement for the values of $x$ where $f(x) < 0$ is $-2 < x < 2$.

The correct choice from the options given is \(\text{$-2 < x < 2$}\).

Therefore, the solution to the problem is $-2 < x < 2$.

First, we examine the provided graph of the quadratic function $f(x)$. The graph clearly shows the x-intercepts (where the function crosses the x-axis) at $x = -3$ and $x = 3$.

Since the quadratic function appears to be a standard parabola opening upwards, the portion of the graph between these two x-intercepts will be above the x-axis, which means that $f(x) > 0$ in this interval.

The intervals to the left of $x = -3$ and to the right of $x = 3$ will be where the graph lies below the x-axis, meaning $f(x) < 0$ in those regions.

Thus, the graph shows that the function $f(x)$ is positive between $x = -3$ and $x = 3$. Therefore, the solution to the problem is:

$-3 < x < 3$