Problem 75 - College Algebra

Question

For the following exercises, evaluate the expressions, given functions \(f, g\), and \(h\) : $$ f(x)=3 x-2 \quad g(x)=5-x^{2} \quad h(x)=-2 x^{2}+3 x-1 $$ $$ f\left(\frac{7}{3}\right)-h(-2) $$

Step-by-Step Solution

Answer

The expression evaluates to 20.

Step 1: Evaluate \( f \left( \frac{7}{3} \right) \)

To find \( f \left( \frac{7}{3} \right) \), substitute \( x = \frac{7}{3} \) into the function \( f(x) = 3x - 2 \). \[ f \left( \frac{7}{3} \right) = 3 \times \frac{7}{3} - 2 \] Simplify the term:\[ = 7 - 2 \] \[ = 5 \] So, \( f \left( \frac{7}{3} \right) = 5 \).

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

To find \( h(-2) \), substitute \( x = -2 \) into the function \( h(x) = -2x^2 + 3x - 1 \).\[ h(-2) = -2(-2)^2 + 3(-2) - 1 \] Calculate each component separately:\[ = -2(4) + (-6) - 1 \] \[ = -8 - 6 - 1 \] Combine the terms:\[ = -15 \] So, \( h(-2) = -15 \).

Step 3: Calculate \( f \left( \frac{7}{3} \right) - h(-2) \)

Subtract the result from Step 2 from the result in Step 1:\[ f \left( \frac{7}{3} \right) - h(-2) = 5 - (-15) \]Rewrite the expression:\[ = 5 + 15 \]Calculate the result:\[ = 20 \]Therefore, the final answer is \( 20 \).

Key Concepts

Algebraic Functions

Algebraic functions are essential in mathematics as they allow us to describe a variety of relationships and operations. These functions can include polynomials, rational functions, and more. Let’s take a closer look at our given functions:

• Function \( f(x) = 3x - 2 \) is a linear function, as it follows the structure of \( ax + b \), where \( a \) and \( b \) are constants. Linear functions are predictable and straightforward to work with.
• Function \( g(x) = 5-x^{2} \) is a quadratic function, denoted by its \( x^{2} \) term. Quadratic functions form parabolas and have important properties like vertex and axis of symmetry.
• Function \( h(x) = -2x^{2} + 3x - 1 \) combines linear and quadratic terms. This makes it a more complex polynomial but still within the realm of algebraic functions.

Understanding these types of functions, recognizing their patterns, and knowing how they behave assist in evaluating expressions effectively.

Substitution Method in Algebra

The substitution method is a useful tool in algebra for simplifying expression evaluation. It involves replacing variables with specific values to solve functions or equations. This approach is particularly beneficial when dealing with function evaluations, as seen in our exercise. Here is how it's applied:

• Substitute \( x = \frac{7}{3} \) in \( f(x) = 3x - 2 \). In simple terms, wherever you see \( x \), you replace it with \( \frac{7}{3} \). This interchange makes the abstract expression more tangible and solvable.
• For \( h(x) = -2x^2 + 3x - 1 \), when substituting \( x = -2 \), each variable in \( h(x) \) becomes \( -2 \). Working on each term separately clarifies the progression and makes arithmetic simpler.

Each step through substitution leads to simpler expressions, making evaluations manageable and clear. By practicing substitution, problems involving varying functions become less daunting.

Arithmetic Operations

Arithmetic operations come in handy after substituting values into functions. They involve performing the basic operations of addition, subtraction, multiplication, and division. This helps in simplifying the expressions even further. Here's how it's applied within our context:

• Once substitution is done, \[ f \left( \frac{7}{3} \right) = 3 \times \frac{7}{3} - 2 \]turns into: \[ 7 - 2 = 5 \]. These steps involve multiplication and subtraction.
• For \( h(-2) \), substitute and simplify:\[ -2(-2)^2 + 3(-2) - 1 \]results as \[ -8 - 6 - 1 = -15 \]. Notice the multiplication, addition, and subtraction here.
• Finally, arithmetic helps in combining results of function evaluations:\[ f \left( \frac{7}{3} \right) - h(-2) = 5 + 15 = 20 \].

Through these operations, mathematical expressions become more understandable and provide the desired outcomes. Mastery of arithmetic opens doors to countless possibilities in solving varied mathematical contexts.