Analyzing $f(x) = (3x^4 + 1)^2$: A Comprehensive Guide

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Welcome to this comprehensive guide on analyzing the function $f(x) = (3x^4 + 1)^2$. In this article, we will delve deep into understanding the characteristics, behavior, and key features of this mathematical expression. Whether you're a student, an educator, or simply someone with a keen interest in mathematics, this guide aims to provide you with a clear and insightful analysis. We'll cover everything from basic properties to more advanced concepts like derivatives and graphical representation. So, let's embark on this mathematical journey together!

Understanding the Basic Properties of $f(x) = (3x^4 + 1)^2$

When we start analyzing any function, it’s crucial to first grasp its basic properties. For $f(x) = (3x^4 + 1)^2$, the function is a polynomial function, specifically a quartic function raised to the power of 2. Let's break down what that means and explore its implications.

Firstly, consider the inner component, $3x^4 + 1$. This is a quartic polynomial because the highest power of $x$ is 4. The coefficient of $x^4$ is 3, which is positive, and the constant term is 1. This means that as $x$ becomes very large (positive or negative), the $3x^4$ term will dominate, and the function will tend towards positive infinity. The addition of 1 simply shifts the graph upwards by one unit.

Now, let’s consider the squaring operation, $(3x^4 + 1)^2$. Squaring any real number results in a non-negative value. Therefore, $f(x)$ will always be greater than or equal to zero. This tells us that the graph of the function will never dip below the x-axis. Additionally, squaring the expression will make the function increase even more rapidly as $x$ moves away from zero. This is because the squaring operation amplifies the already large values produced by the quartic term.

Another crucial property to consider is the symmetry of the function. Since $x$ is raised to an even power (4), the function is an even function. This means that $f(x) = f(-x)$ for all $x$. Graphically, this implies that the function is symmetric about the y-axis. If you were to fold the graph along the y-axis, the two halves would perfectly overlap. This symmetry simplifies the analysis because we only need to focus on the behavior of the function for $x ">= 0$, and we can infer the behavior for $x < 0$.

In summary, understanding these basic properties – the quartic nature of the inner polynomial, the squaring operation ensuring non-negativity, and the even symmetry – provides a strong foundation for further analysis. It allows us to anticipate the function's overall shape and behavior before even delving into more complex calculus concepts.

Determining the Domain and Range of $f(x)$

Understanding the domain and range of a function is essential for a complete analysis. The domain represents all possible input values (x-values) for which the function is defined, while the range represents all possible output values (y-values) that the function can produce. Let’s determine these for $f(x) = (3x^4 + 1)^2$.

For the domain, we need to consider if there are any restrictions on the values of $x$ that can be plugged into the function. In this case, $f(x)$ is a polynomial function, and polynomial functions are defined for all real numbers. There are no denominators that could be zero, no square roots of negative numbers, or any other restrictions that would limit the possible input values. Therefore, the domain of $f(x)$ is all real numbers, which can be written as $(-\infty, \infty)$.

Now, let's consider the range. We know from our earlier analysis that $f(x)$ is always non-negative due to the squaring operation. The minimum value of the function occurs when the inner expression, $3x^4 + 1$, is at its minimum. Since $x^4$ is always non-negative, the minimum value of $3x^4$ is 0, which occurs when $x = 0$. Therefore, the minimum value of $3x^4 + 1$ is $3(0)^4 + 1 = 1$. When we square this, we get $f(0) = (1)^2 = 1$. This means that the lowest point on the graph of the function is at $y = 1$.

As $x$ moves away from 0 in either direction (positive or negative), the term $3x^4$ increases rapidly, and so does $3x^4 + 1$. Squaring this expression further amplifies the growth. As $x$ approaches positive or negative infinity, $f(x)$ also approaches positive infinity. Therefore, the function can take on any value greater than or equal to 1.

In conclusion, the range of $f(x)$ is all real numbers greater than or equal to 1, which can be written in interval notation as $[1, \infty)$. Knowing the domain and range helps us visualize the function's scope and potential behavior, setting the stage for more detailed investigations.

Finding Intercepts of the Function

Intercepts are the points where the graph of a function intersects the coordinate axes. Finding the intercepts is a crucial step in understanding the graph's position and orientation in the coordinate plane. There are two types of intercepts we typically look for: the y-intercept and the x-intercept(s). Let's find them for $f(x) = (3x^4 + 1)^2$.

The y-intercept is the point where the graph intersects the y-axis. This occurs when $x = 0$. To find the y-intercept, we substitute $x = 0$ into the function: $f(0) = (3(0)^4 + 1)^2 = (0 + 1)^2 = 1^2 = 1$. So, the y-intercept is the point $(0, 1)$. This point represents the function's value when the input is zero, giving us a crucial reference point on the graph.

Next, we look for the x-intercepts. These are the points where the graph intersects the x-axis, which occur when $f(x) = 0$. To find the x-intercepts, we set the function equal to zero and solve for $x$: $(3x^4 + 1)^2 = 0$. Taking the square root of both sides gives us $3x^4 + 1 = 0$. Now, we need to solve for $x$. Subtracting 1 from both sides gives $3x^4 = -1$. Dividing by 3 gives $x^4 = -\frac{1}{3}$.

Here's where we encounter an important realization. Since $x^4$ is always non-negative for real values of $x$, there is no real number $x$ that will satisfy the equation $x^4 = -\frac{1}{3}$. A real number raised to an even power cannot be negative. Therefore, there are no real x-intercepts for this function. This tells us that the graph of $f(x)$ never crosses the x-axis, which is consistent with our earlier finding that the range of the function is $[1, \infty)$.

In summary, the function $f(x) = (3x^4 + 1)^2$ has one y-intercept at $(0, 1)$ and no real x-intercepts. This information helps us sketch a more accurate graph of the function and understand its behavior near the axes.

Analyzing the First Derivative: Increasing and Decreasing Intervals

The first derivative of a function, denoted as $f'(x)$, provides valuable insights into the function’s rate of change. Specifically, it helps us determine where the function is increasing, decreasing, or stationary (i.e., has a slope of zero). Let’s calculate the first derivative of $f(x) = (3x^4 + 1)^2$ and use it to analyze the increasing and decreasing intervals.

First, we need to find the first derivative $f'(x)$. We’ll use the chain rule, which states that if we have a composite function $f(g(x))$, then its derivative is $f'(g(x)) \cdot g'(x)$. In our case, we can think of $f(x)$ as $(u)^2$, where $u = 3x^4 + 1$. So, the derivative will be:

$f'(x) = 2(3x^4 + 1) \cdot \frac{d}{dx}(3x^4 + 1)$

Now, we need to find the derivative of the inner function $3x^4 + 1$. Using the power rule, which states that $\frac{d}{dx}(x^n) = nx^{n-1}$, we get:

$\frac{d}{dx}(3x^4 + 1) = 12x^3$

Substituting this back into our expression for $f'(x)$, we have:

$f'(x) = 2(3x^4 + 1)(12x^3) = 24x^3(3x^4 + 1)$

Now that we have the first derivative, $f'(x) = 24x^3(3x^4 + 1)$, we can analyze its sign to determine the intervals of increasing and decreasing behavior. A function is increasing where its first derivative is positive ($f'(x) > 0$), decreasing where its first derivative is negative ($f'(x) < 0$), and stationary where its first derivative is zero ($f'(x) = 0$).

To find the critical points, we set $f'(x) = 0$ and solve for $x$:

$24x^3(3x^4 + 1) = 0$

This equation is satisfied if either $24x^3 = 0$ or $3x^4 + 1 = 0$. The second part, $3x^4 + 1 = 0$, has no real solutions (as we discussed earlier when finding the x-intercepts). The first part, $24x^3 = 0$, gives us $x^3 = 0$, which means $x = 0$ is the only critical point.

Now, we can create a sign chart for $f'(x)$ using the critical point $x = 0$. We'll test values in the intervals $(-\infty, 0)$ and $(0, \infty)$ to determine the sign of $f'(x)$ in each interval:

• For $x < 0$ (e.g., $x = -1$), $f'(-1) = 24(-1)^3(3(-1)^4 + 1) = 24(-1)(3 + 1) = -96$, which is negative. Therefore, $f(x)$ is decreasing on the interval $(-\infty, 0)$.
• For $x > 0$ (e.g., $x = 1$), $f'(1) = 24(1)^3(3(1)^4 + 1) = 24(1)(3 + 1) = 96$, which is positive. Therefore, $f(x)$ is increasing on the interval $(0, \infty)$.

In conclusion, by analyzing the first derivative, we have determined that $f(x) = (3x^4 + 1)^2$ is decreasing on the interval $(-\infty, 0)$ and increasing on the interval $(0, \infty)$. The critical point at $x = 0$ is a local minimum, which we will explore further in the next section.

Analyzing the Second Derivative: Concavity and Inflection Points

The second derivative, denoted as $f''(x)$, provides information about the concavity of a function. Concavity describes the curvature of the graph; a function is concave up where its graph is curved upwards (like a smile) and concave down where its graph is curved downwards (like a frown). Inflection points are the points where the concavity changes. Let's analyze the second derivative of $f(x) = (3x^4 + 1)^2$ to determine the concavity and inflection points.

First, we need to find the second derivative, $f''(x)$. We already found the first derivative to be $f'(x) = 24x^3(3x^4 + 1)$. To find the second derivative, we need to differentiate $f'(x)$ with respect to $x$. We'll use the product rule, which states that if we have a product of two functions, $(uv)' = u'v + uv'$. In this case, let $u = 24x^3$ and $v = 3x^4 + 1$.

The derivatives of $u$ and $v$ are:

• $u' = \frac{d}{dx}(24x^3) = 72x^2$
• $v' = \frac{d}{dx}(3x^4 + 1) = 12x^3$

Applying the product rule, we get:

$f''(x) = u'v + uv' = (72x^2)(3x^4 + 1) + (24x^3)(12x^3)$

Now, we simplify the expression:

$f''(x) = 216x^6 + 72x^2 + 288x^6 = 504x^6 + 72x^2$

We can factor out $72x^2$ from the expression:

$f''(x) = 72x^2(7x^4 + 1)$

To determine the intervals of concavity and inflection points, we need to analyze the sign of $f''(x)$. A function is concave up where $f''(x) > 0$, concave down where $f''(x) < 0$, and an inflection point occurs where $f''(x)$ changes sign.

First, we find the potential inflection points by setting $f''(x) = 0$ and solving for $x$:

$72x^2(7x^4 + 1) = 0$

This equation is satisfied if either $72x^2 = 0$ or $7x^4 + 1 = 0$. The second part, $7x^4 + 1 = 0$, has no real solutions because $7x^4$ is always non-negative, so $7x^4 + 1$ will always be greater than or equal to 1. The first part, $72x^2 = 0$, gives us $x^2 = 0$, which means $x = 0$ is a potential inflection point.

Now, we create a sign chart for $f''(x)$ using the potential inflection point $x = 0$. However, since $x^2$ is always non-negative and $7x^4 + 1$ is always positive, their product $72x^2(7x^4 + 1)$ will also always be non-negative. This means that $f''(x)$ is always greater than or equal to zero.

• For $x \neq 0$, $f''(x) > 0$, so the function is concave up.
• At $x = 0$, $f''(0) = 0$, but the concavity does not change because $f''(x)$ is non-negative on both sides of $x = 0$. Therefore, $x = 0$ is not an inflection point.

In conclusion, the function $f(x) = (3x^4 + 1)^2$ is concave up for all $x \neq 0$ and has no inflection points. This indicates that the graph of the function is always curved upwards.

Sketching the Graph of $f(x) = (3x^4 + 1)^2$

Now that we have gathered a wealth of information about the function $f(x) = (3x^4 + 1)^2$, we can put it all together to sketch an accurate graph. We know the following:

• Basic Properties: It's an even function, symmetric about the y-axis.
• Domain: $(-\infty, \infty)$
• Range: $[1, \infty)$
• Intercepts: y-intercept at $(0, 1)$, no x-intercepts
• Increasing/Decreasing Intervals: Decreasing on $(-\infty, 0)$, increasing on $(0, \infty)$
• Local Minimum: At $x = 0$, with a value of $f(0) = 1$
• Concavity: Concave up for all $x \neq 0$
• Inflection Points: None

With this information, we can create a sketch of the graph:

1. Start with the symmetry: Since the function is even, we can focus on the positive x-axis and then reflect the graph across the y-axis.
2. Plot the y-intercept: The graph passes through $(0, 1)$. This is also the minimum point of the function.
3. Consider the increasing and decreasing intervals: The function decreases as we move from left to right until we reach $x = 0$, and then it increases as we move to the right.
4. Account for concavity: The function is always concave up, so the graph will always be curved upwards.
5. Reflect and extend: Reflect the portion of the graph for $x \geq 0$ across the y-axis to complete the graph for all $x$. The graph extends upwards towards infinity as $x$ goes to positive or negative infinity.

The resulting graph is a U-shaped curve, symmetric about the y-axis, with its lowest point at $(0, 1)$. It never touches the x-axis and is always curved upwards.

In summary, by synthesizing all the information we gathered from analyzing the function, its derivatives, intercepts, and domain/range, we can create an accurate sketch of the graph of $f(x) = (3x^4 + 1)^2$. This visual representation provides a powerful tool for understanding the function’s behavior and characteristics.

Conclusion

In this comprehensive guide, we have thoroughly analyzed the function $f(x) = (3x^4 + 1)^2$. We explored its basic properties, determined its domain and range, found its intercepts, analyzed its first and second derivatives to identify increasing and decreasing intervals, local minima, concavity, and inflection points. Finally, we synthesized all this information to sketch an accurate graph of the function.

Understanding the behavior of functions is a fundamental skill in mathematics, and this analysis provides a detailed example of how to approach such problems. By breaking down the function into its key components and using calculus techniques, we can gain a deep understanding of its characteristics and behavior.

For further exploration of mathematical functions and their analysis, you might find resources like Khan Academy's Calculus section helpful. Keep practicing and exploring, and you'll continue to develop your mathematical analysis skills!