Evaluating H(x) = X^4 - 4x^2 + 1 For Different Values

by ADMIN 54 views

This article delves into the process of evaluating the function h(x) = x^4 - 4x^2 + 1 for different values of the independent variable x. We will explore how to substitute various numerical and algebraic expressions into the function and simplify the results. This exercise is crucial for understanding function behavior and lays the groundwork for more advanced calculus concepts. We will tackle specific examples, including evaluating h(x) at x = -3, x = -1, x = -x, and x = 3a. By working through these examples step-by-step, you'll gain a solid understanding of function evaluation and algebraic manipulation.

a. Evaluating h(-3)

To evaluate h(-3), we substitute -3 for x in the function h(x) = x^4 - 4x^2 + 1. This process involves replacing every instance of x with -3 and then simplifying the resulting expression. Careful attention to the order of operations (PEMDAS/BODMAS) is essential to arrive at the correct answer. This substitution allows us to determine the value of the function when the input is -3, providing a specific point on the graph of the function. Function evaluation is a fundamental skill in mathematics, allowing us to understand how a function transforms different inputs into corresponding outputs. By understanding this process, we can predict the function's behavior and its relationship to the input variable. This also helps in visualizing the function's graph and identifying key features like intercepts and extrema. When we substitute -3 into the function, we are essentially asking, "What is the output of the function when the input is -3?" The subsequent simplification will provide the answer. This simple question is the cornerstone of function evaluation, and mastering it is critical for success in higher-level mathematics. Evaluating functions at specific points is also vital in various applications, such as modeling physical phenomena or analyzing data. The result of h(-3) will be a numerical value that represents the function's output at that particular input.

Here's the step-by-step evaluation:

  • h(-3) = (-3)^4 - 4(-3)^2 + 1
  • h(-3) = 81 - 4(9) + 1
  • h(-3) = 81 - 36 + 1
  • h(-3) = 45 + 1
  • h(-3) = 46

Therefore, the value of the function h(x) when x is -3 is 46. This means that the point (-3, 46) lies on the graph of the function. This single evaluation provides valuable information about the function's behavior in the vicinity of x = -3. Further evaluations at other points can paint a more complete picture of the function's overall shape and characteristics. The ability to accurately evaluate functions is a prerequisite for more advanced tasks, such as finding roots, determining intervals of increase and decrease, and sketching graphs. The value 46 represents the y-coordinate corresponding to the x-coordinate of -3 on the function's graph. It is a specific output of the function for a given input, and such evaluations are crucial for understanding the function's mapping behavior.

b. Evaluating h(-1)

Similarly, to evaluate h(-1), we substitute -1 for x in the function h(x) = x^4 - 4x^2 + 1. This substitution is another example of how we determine the function's output for a specific input value. The process remains the same: replace every instance of x with -1 and then simplify the expression using the correct order of operations. This evaluation will give us another point on the function's graph, contributing to our understanding of its overall shape. Understanding function evaluation is crucial because it's the fundamental way we extract information from a function. By plugging in different input values, we can observe how the function transforms those inputs and generates corresponding outputs. This process is essential for analyzing the function's behavior, identifying its key features, and applying it to real-world problems. For example, in a physics context, h(x) might represent the height of a projectile at time x, and evaluating h(-1) would tell us the height at time -1 (although negative time might not have a physical interpretation in this scenario, the mathematical process remains the same). The result of h(-1) is a single numerical value that tells us the function's output when the input is -1. This value is a critical piece of information for understanding the function's behavior.

Here's the step-by-step evaluation:

  • h(-1) = (-1)^4 - 4(-1)^2 + 1
  • h(-1) = 1 - 4(1) + 1
  • h(-1) = 1 - 4 + 1
  • h(-1) = -3 + 1
  • h(-1) = -2

Thus, the value of the function h(x) when x is -1 is -2. This indicates that the point (-1, -2) lies on the graph of the function. Comparing this result with the previous evaluation of h(-3), we begin to see how the function's output changes as the input changes. The negative result of -2 for h(-1) suggests that the function's graph is below the x-axis at x = -1. This type of information is invaluable when sketching the graph of a function and understanding its behavior. Furthermore, knowing that h(-1) = -2 allows us to compare the function's value at different points, which can help in identifying trends and patterns. This specific evaluation contributes to the overall understanding of the function's properties and characteristics. By evaluating at different points, we begin to piece together the function's behavior like assembling pieces of a puzzle.

c. Evaluating h(-x)

Evaluating h(-x) involves substituting -x for x in the function h(x) = x^4 - 4x^2 + 1. This is a slightly different type of evaluation compared to the previous numerical examples. Here, we are substituting an algebraic expression rather than a specific number. The process, however, remains the same: replace every x with -x and simplify. This type of evaluation is crucial for understanding function symmetry and identifying whether a function is even or odd. An even function satisfies the condition f(-x) = f(x), while an odd function satisfies f(-x) = -f(x). Evaluating h(-x) will help us determine if our function h(x) exhibits either of these symmetries. Understanding function symmetry simplifies graphing and analysis, as it reveals patterns in the function's behavior. If h(-x) turns out to be equal to h(x), then we know the function is symmetric about the y-axis. This means that the left half of the graph is a mirror image of the right half. Such symmetries are valuable shortcuts in understanding and visualizing functions. Evaluating h(-x) also provides insights into how the function transforms negative inputs in relation to positive inputs. It allows us to see the algebraic effect of changing the sign of the input variable. This is a critical skill in function analysis and manipulation.

Here's the step-by-step evaluation:

  • h(-x) = (-x)^4 - 4(-x)^2 + 1
  • h(-x) = x^4 - 4x^2 + 1

Notice that (-x)^4 = x^4 because a negative number raised to an even power is positive, and (-x)^2 = x^2 for the same reason.

  • h(-x) = x^4 - 4x^2 + 1

Comparing this result to the original function h(x) = x^4 - 4x^2 + 1, we observe that h(-x) = h(x). This confirms that the function h(x) is an even function. This finding has significant implications for the function's graph, as it implies symmetry about the y-axis. Knowing this symmetry allows us to sketch the graph more easily, as we only need to analyze the function's behavior for positive x values and then reflect that behavior across the y-axis. The fact that h(-x) = h(x) is a powerful piece of information that simplifies the analysis and visualization of the function. This evaluation also demonstrates the importance of understanding the properties of exponents and how they affect the sign of a term. The result reinforces the concept of even functions and their characteristic symmetry.

d. Evaluating h(3a)

Finally, to evaluate h(3a), we substitute 3a for x in the function h(x) = x^4 - 4x^2 + 1. This evaluation involves substituting an algebraic expression containing a constant (a) for the independent variable x. The process remains consistent: replace each instance of x with 3a and simplify the resulting expression. This type of evaluation is helpful in understanding how the function behaves with scaled inputs. The constant a acts as a scaling factor, and evaluating h(3a) allows us to see how this scaling affects the function's output. This skill is valuable in various contexts, such as transformations of graphs and solving equations involving functions. When dealing with algebraic substitutions, it's crucial to pay attention to the order of operations and to correctly apply the rules of exponents. The result of h(3a) will be an algebraic expression in terms of a, representing the function's output when the input is 3a. This expression can then be further analyzed or used in other calculations. Evaluating functions with algebraic inputs is a key step in understanding their general behavior and how they respond to different types of transformations.

Here's the step-by-step evaluation:

  • h(3a) = (3a)^4 - 4(3a)^2 + 1
  • h(3a) = 81a^4 - 4(9a^2) + 1
  • h(3a) = 81a^4 - 36a^2 + 1

The result, h(3a) = 81a^4 - 36a^2 + 1, is a polynomial expression in terms of a. This expression represents the function's output when the input is 3a. The presence of a^4 and a^2 terms indicates that the function's behavior with the scaled input 3a will be influenced by the value of a. This evaluation demonstrates how functions can transform algebraic inputs into new algebraic expressions. The resulting polynomial can be further analyzed, for example, by finding its roots or determining its extrema. The ability to evaluate functions with algebraic inputs is essential for many areas of mathematics, including calculus and differential equations. This particular evaluation highlights the effect of scaling the input variable by a constant factor. The coefficients in the resulting polynomial reflect the scaling effect on the original function's terms.

In conclusion, we have successfully evaluated the function h(x) = x^4 - 4x^2 + 1 for various inputs, including numerical values and algebraic expressions. These evaluations have provided insights into the function's behavior and have demonstrated the fundamental process of function evaluation. We found that h(-3) = 46, h(-1) = -2, h(-x) = h(x) (indicating that h(x) is an even function), and h(3a) = 81a^4 - 36a^2 + 1. These examples showcase the versatility of function evaluation and its importance in understanding and analyzing mathematical functions.