Problem 32 - College Algebra Essentials

Question

Evaluate each function at the given values of the independent variable and simplify. $$ h(x)=x^{3}-x+1 $$ a. \(h(3)\) b. \(h(-2)\) c. \(h(-x)\) d. \(h(3 a)\)

Step-by-Step Solution

Answer

The simplified answers are:\na. \(h(3) = 25\)\nb. \(h(-2) = -5\)\nc. \(h(-x) = -x^{3} + x + 1\)\nd. \(h(3a) = 27a^{3} - 3a + 1\).

Step 1: Evaluate \(h(3)\)

Substitute \(x\) with 3 in the function \(h(x)\). It becomes \(h(3) = 3^{3} - 3 + 1 = 27 - 3 + 1 = 25\)

Step 2: Evaluate \(h(-2)\)

Substitute \(x\) with -2 in the function \(h(x)\). It becomes \(h(-2) = (-2)^{3} - (-2) + 1 = -8 + 2 + 1 = -5\)

Step 3: Evaluate \(h(-x)\)

Substitute \(x\) with -x in the function \(h(x)\). It becomes \(h(-x) = (-x)^{3} - (-x) + 1 = -x^{3} + x + 1\)

Step 4: Evaluate \(h(3a)\)

Substitute \(x\) with \(3a\) in the function \(h(x)\). It becomes \(h(3a) = (3a)^{3} - 3a + 1 = 27a^{3} - 3a + 1\).

Key Concepts

Polynomial Functions

Polynomial functions are one of the most fundamental types of functions used in algebra and are based on powers of an independent variable. A polynomial function like the one in the exercise, h(x) = x^3 - x + 1, consists of terms with variables raised to non-negative integer exponents.

In this example, h(x) is a third-degree polynomial because the highest power of the variable x is three. The coefficients of the terms here are 1 for x^3, -1 for x, and 1 for the constant term. Polynomial functions often have various x-intercepts and extrema, which makes them interesting subjects of study in college algebra.

Independent Variable Substitution

Independent variable substitution is a technique used to evaluate functions for specific values. This is crucial when finding the output of a function given different inputs. In our example with the function h(x) = x^3 - x + 1, we replace x with the given number or expression to find the function's value at that point. For instance, solving the function h(3) involves substituting x with 3 and computing the resulting expression.

Algebraic Expressions Simplification

Algebraic expressions simplification involves reducing an expression to its simplest form. This often means applying arithmetic operations, combining like terms, and using algebraic properties such as the distributive property to condense an expression. For the function h(x), when we evaluate it for different values, we pursue a simplified form; for example, h(3) = 3^3 - 3 + 1 simplifies to 25 after we perform the necessary calculations. Simplifying expressions is a key skill in algebra that helps with understanding the behavior of functions and solving equations more easily.

College Algebra

College Algebra encompasses the study of algebraic concepts beyond the high school level and is foundational for various areas of mathematics and science. It includes an array of topics from polynomial functions, equation systems, to complex numbers, and their properties. In the context of our exercise, college algebra includes understanding how functions work and how to manipulate algebraic expressions, enabling us to evaluate polynomial functions like h(x) at different values of the independent variable effectively. This subject lays the groundwork for more advanced fields such as calculus and linear algebra.