Problem 54 - Intermediate Algebra

Question

If \(f(x)=2 x^{2}-7\) and \(g(x)=x^{2}+x-1\), find \(f(-2)\), \(f(3), g(-4)\), and \(g(5)\).

Step-by-Step Solution

Answer

\(f(-2) = 1\), \(f(3) = 11\); \(g(-4) = 11\), \(g(5) = 29\).

Step 1: Evaluate f(-2)

To find \(f(-2)\), substitute \(-2\) for \(x\) in the function \(f(x)=2x^2-7\). This gives: \[f(-2) = 2(-2)^2 - 7.\] Calculate \(-2^2 = 4\), then \(2\cdot 4 = 8\). So, \[f(-2) = 8 - 7 = 1.\]

Step 2: Evaluate f(3)

Substitute \(3\) for \(x\) in \(f(x)=2x^2-7\): \[f(3) = 2(3)^2 - 7.\] Calculate \(3^2 = 9\), then \(2 \cdot 9 = 18\). So, \[f(3) = 18 - 7 = 11.\]

Step 3: Evaluate g(-4)

Substitute \(-4\) for \(x\) in \(g(x)=x^2+x-1\): \[g(-4) = (-4)^2 + (-4) - 1.\] Calculate \(-4^2 = 16\), \(16 + (-4) = 12\), and finally \(12 - 1 = 11\). Therefore, \[g(-4) = 11.\]

Step 4: Evaluate g(5)

Substitute \(5\) for \(x\) in \(g(x)=x^2+x-1\): \[g(5) = (5)^2 + 5 - 1.\] Calculate \(5^2 = 25\), \(25 + 5 = 30\), and then \(30 - 1 = 29\). Thus, \[g(5) = 29.\]

Key Concepts

Polynomial Functions

Polynomial functions are algebraic expressions that involve variables raised to whole number powers, multiplied by coefficients. They form the basis of many mathematical models, and understanding their structure is crucial for solving various real-world problems.
Consider the function \( f(x) = 2x^2 - 7 \). Here, the expression is a polynomial because it consists of terms involving powers of the variable \(x\). The term \(2x^2\) is the quadratic term, and \(-7\) is the constant term. The degree of a polynomial is given by the highest power of \(x\), which in this case is 2.
A polynomial can have multiple terms, each contributing to its overall shape when graphed. The function \( g(x) = x^2 + x - 1 \) is also a polynomial, with the highest degree term \(x^2\), indicating it is a quadratic polynomial. Understanding these terms allows you to determine the behavior and graph of the polynomial.

Substitution Method

The substitution method is used to evaluate functions by replacing the variable with a specific number. This technique simplifies the expression, so you can easily calculate the result.

For example, to find \( f(-2) \) for the function \( f(x) = 2x^2 - 7 \), you substitute \(-2\) for \(x\):

• Calculate \( (-2)^2 \), which equals 4.
• Multiply 4 by 2 to get 8.
• Finally, subtract 7 to get 1.

The process repeats for each value you wish to evaluate the function at, transforming an algebraic expression into a simple arithmetic calculation.
This method is particularly useful for checking the value of a polynomial function at different points and is essential for graphing or analyzing functions.

Algebraic Expressions

Algebraic expressions are combinations of numbers and variables linked by operations like addition, subtraction, multiplication, and division. They form the backbone of algebra and appear in all sorts of mathematical problems.
In the context of polynomial functions, like \( f(x) = 2x^2 - 7 \) and \( g(x) = x^2 + x - 1 \), algebraic expressions are present in each term. These functions are made up of simpler expressions such as \( 2x^2 \), \(-7\), and \( x^2 + x \), which are combined to form the whole expression.
Breaking down complex algebraic expressions into smaller parts can simplify the evaluation or manipulation of the function. Understanding how these components interact is crucial in solving equations and interpreting mathematical information, making skills in manipulating algebraic expressions vital for any mathematics student.