Evaluating H(x) = X^4 + 2x^2 + 3 For Various Inputs
In this article, we will delve into the evaluation of the function h(x) = x^4 + 2x^2 + 3 at various values of the independent variable x. This exercise is fundamental in understanding function behavior and is a crucial skill in algebra and calculus. We will systematically evaluate the function for the given inputs, simplifying the results along the way. This comprehensive guide aims to provide a clear and concise explanation of each step, ensuring that readers can confidently apply these techniques to other functions and problems.
a. Evaluating h(-3)
To evaluate h(-3), we substitute x with -3 in the function h(x) = x^4 + 2x^2 + 3. This means we replace every instance of x in the function's expression with -3. The calculation proceeds as follows:
h(-3) = (-3)^4 + 2(-3)^2 + 3
First, we calculate (-3)^4, which means -3 multiplied by itself four times: (-3) * (-3) * (-3) * (-3). Since a negative number raised to an even power results in a positive number, (-3)^4 equals 81.
Next, we calculate (-3)^2, which is -3 multiplied by itself: (-3) * (-3). This results in 9. We then multiply this result by 2, giving us 2 * 9 = 18.
Now, we substitute these values back into the expression:
h(-3) = 81 + 18 + 3
Finally, we add the numbers together: 81 + 18 + 3 = 102. Therefore,
h(-3) = 102
This result tells us that the function h(x) outputs the value 102 when the input is -3. Understanding this process of substitution and simplification is crucial for working with functions in mathematics. This example showcases how a seemingly complex expression can be broken down into simpler steps to arrive at a clear and accurate answer. The ability to evaluate functions at specific points is essential for graphing functions, solving equations, and understanding various mathematical concepts.
b. Evaluating h(-1)
Now, let's evaluate h(-1). Similar to the previous example, we substitute x with -1 in the function h(x) = x^4 + 2x^2 + 3. This process involves replacing every instance of x in the function's expression with -1. The calculation unfolds as follows:
h(-1) = (-1)^4 + 2(-1)^2 + 3
First, we calculate (-1)^4, which means -1 multiplied by itself four times: (-1) * (-1) * (-1) * (-1). As before, a negative number raised to an even power yields a positive number, so (-1)^4 equals 1.
Next, we calculate (-1)^2, which is -1 multiplied by itself: (-1) * (-1). This results in 1. We then multiply this result by 2, giving us 2 * 1 = 2.
Substituting these values back into the expression, we get:
h(-1) = 1 + 2 + 3
Finally, we add the numbers together: 1 + 2 + 3 = 6. Therefore,
h(-1) = 6
This result indicates that the function h(x) outputs the value 6 when the input is -1. This evaluation demonstrates the function's behavior at a different point and further reinforces the process of substitution and simplification. It's important to note how the even powers in the function's expression result in positive values, regardless of the sign of the input. This characteristic is crucial in understanding the overall shape and properties of the function's graph. By evaluating functions at various points, we gain a better understanding of their behavior and characteristics.
c. Evaluating h(-x)
Evaluating h(-x) involves substituting x with -x in the function h(x) = x^4 + 2x^2 + 3. This substitution allows us to analyze the function's symmetry and behavior with respect to the x-axis. The calculation proceeds as follows:
h(-x) = (-x)^4 + 2(-x)^2 + 3
First, we calculate (-x)^4. This means -x multiplied by itself four times: (-x) * (-x) * (-x) * (-x). Since a negative value raised to an even power results in a positive value, (-x)^4 equals x^4.
Next, we calculate (-x)^2, which is -x multiplied by itself: (-x) * (-x). This results in x^2. We then multiply this result by 2, giving us 2 * x^2.
Substituting these values back into the expression, we get:
h(-x) = x^4 + 2x^2 + 3
Notice that this is the same as the original function, h(x). Therefore,
h(-x) = h(x)
This result indicates that the function h(x) is an even function. An even function is defined as a function that satisfies the condition f(-x) = f(x) for all x in its domain. This property implies that the graph of the function is symmetric with respect to the y-axis. In other words, the left and right sides of the graph are mirror images of each other. Understanding the concept of even functions is crucial in various areas of mathematics, including calculus and differential equations. This evaluation demonstrates how substituting -x into a function can reveal its symmetry properties.
d. Evaluating h(3a)
Finally, let's evaluate h(3a). In this case, we substitute x with 3a in the function h(x) = x^4 + 2x^2 + 3. This substitution allows us to express the function's output in terms of the variable a. The calculation unfolds as follows:
h(3a) = (3a)^4 + 2(3a)^2 + 3
First, we calculate (3a)^4. This means 3a multiplied by itself four times: (3a) * (3a) * (3a) * (3a). Using the properties of exponents, we can rewrite this as 3^4 * a^4, which equals 81a^4.
Next, we calculate (3a)^2, which is 3a multiplied by itself: (3a) * (3a). This results in 9a^2. We then multiply this result by 2, giving us 2 * 9a^2 = 18a^2.
Substituting these values back into the expression, we get:
h(3a) = 81a^4 + 18a^2 + 3
Therefore,
h(3a) = 81a^4 + 18a^2 + 3
This result expresses the function's output in terms of the variable a. This type of evaluation is essential when working with composite functions or when analyzing the function's behavior for a range of values. The expression 81a^4 + 18a^2 + 3 represents a polynomial in a, and its properties can be further analyzed using various algebraic techniques. This example showcases how substituting a variable expression into a function can lead to a new expression that reveals different aspects of the function's behavior.
In conclusion, we have successfully evaluated the function h(x) = x^4 + 2x^2 + 3 at the given values of the independent variable. These examples demonstrate the fundamental process of substitution and simplification, which is crucial for working with functions in mathematics. By understanding how to evaluate functions at specific points and expressions, we gain a deeper understanding of their behavior and properties. This knowledge is essential for various applications in algebra, calculus, and other areas of mathematics.
