Problem 25 - College Algebra

Question

In Exercises \(21-32,\) evaluate each function at the given values of the independent variable and simplify. $$h(x)=x^{4}-x^{2}+1$$ a. \(h(2)\) b. \(h(-1)\) c. \(h(-x)\) d. \(h(3 a)\)

Step-by-Step Solution

Answer

h(2) = 13, h(-1) = 1, h(-x) = x^4 - x^2 + 1, h(3a) = 81a^4 - 9a^2 + 1

Step 1: Evaluate h(2)

Substitute \(2\) for \(x\) in \(h(x)\) to get \(h(2) = 2^4 - 2^2 + 1 = 16 - 4 + 1 = 13\)

Step 2: Evaluate h(-1)

Substitute \(-1\) for \(x\) in \(h(x)\) to get \(h(-1) = (-1)^4 - (-1)^2 + 1 = 1 - 1 + 1 = 1\)

Step 3: Evaluate h(-x)

Substitute \(-x\) for \(x\) in \(h(x)\) to get \(h(-x) = (-x)^4 - (-x)^2 + 1 = x^4 - x^2 + 1\)

Step 4: Evaluate h(3a)

Substitute \(3a\) for \(x\) in \(h(x)\) to get \(h(3a) = (3a)^4 - (3a)^2 + 1 = 81a^4 - 9a^2 + 1\)

Key Concepts

Function Evaluation

When we talk about function evaluation, we are essentially discussing how to find the value of a function for a specific variable input. In practical terms, it involves replacing the variable in the function's expression with a given number or expression, and then performing the calculations.

In the original exercise, we evaluated the function \( h(x) = x^{4} - x^{2} + 1 \) at various points. Let's explore these steps to better grasp the process:

• For \( h(2) \): Replace \( x \) with \( 2 \), giving \( 2^4 - 2^2 + 1 = 13 \). This is the value of the function at \( x = 2 \).
• For \( h(-1) \): Substitute \( x \) with \( -1 \), resulting in \( (-1)^4 - (-1)^2 + 1 = 1 \). Negative inputs can sometimes change the outcome!
• For \( h(-x) \): Use \( -x \) instead of \( x \), which gives \( (-x)^4 - (-x)^2 + 1 = x^4 - x^2 + 1 \). Notice how the expression is the same as the original function.
• For \( h(3a) \): Input \( 3a \) for \( x \), yielding \( (3a)^4 - (3a)^2 + 1 = 81a^4 - 9a^2 + 1 \). Here, we see how variables can transform a function expression dramatically.

Understanding function evaluation allows us to determine specific outcomes based on different variables, providing a deeper insight into how the function behaves.

Exponents

Exponents are a fundamental part of algebra and they tell us how many times to multiply the base number by itself. For example, \( x^4 \) means \( x \times x \times x \times x \). They are crucial when dealing with polynomial functions, like the one in the original exercise.

In the problem, we see several instances of exponents:

• For \( h(x) = x^4 - x^2 + 1 \), the exponent \( 4 \) tells us that \( x \) is multiplied by itself four times.
• When evaluating \( h(2) \) and calculating \( 2^4 \), it leads to \( 16 \), since \( 2 \times 2 \times 2 \times 2 = 16 \).
• Additionally, \( 2^2 \) reduces to \( 4 \) because \( 2 \times 2 = 4 \).
• For \( (-1)^4 \), despite being a negative base, the result is \( 1 \) because the even exponent flips the negative sign.

Exponents can greatly influence the value of polynomial functions, making them essential for understanding the behavior and outputs of complex expressions.

Function Simplification

Function simplification involves reducing a mathematical expression to its simplest form. This means performing all operations, combining like terms, and sometimes factoring, if necessary, to make expressions more understandable and manageable.

In our exercise, each step of evaluating \( h(x) \) also included simplification. Let's underscore these moments:

• When finding \( h(2) = 16 - 4 + 1 \), the operands were combined to yield the simplified result, \( 13 \).
• In evaluating \( h(-x) = x^4 - x^2 + 1 \), simplification was crucial to show that the expression remains unchanged since all terms were already simplified.
• For \( h(3a) = 81a^4 - 9a^2 + 1 \), multiplication and simplification of terms were performed quickly after determining \( (3a)^4 \) as \( 81a^4 \) and \( (3a)^2 \) as \( 9a^2 \).

Understanding simplification helps in handling complex expressions efficiently, making subsequent calculations and evaluations more straightforward.