Problem 41 - Algebra 2

Question

If \(f(x)=4 x, g(x)=2 x-1,\) and \(h(x)=x^{2}+1,\) find each value. $$ g\left[h\left(-\frac{1}{2}\right)\right] $$

Step-by-Step Solution

Answer

The value is \( \frac{3}{2} \).

Step 1: Evaluate \( h(-\frac{1}{2}) \)

To find \( h(-\frac{1}{2}) \), substitute \(-\frac{1}{2}\) into the function \( h(x) = x^2 + 1 \). Calculate the expression: \[h(-\frac{1}{2}) = \left(-\frac{1}{2}\right)^2 + 1 = \frac{1}{4} + 1 = \frac{5}{4}\]

Step 2: Evaluate \( g(h(-\frac{1}{2})) \)

Now use the result from Step 1, \( h(-\frac{1}{2}) = \frac{5}{4} \), as the input for \( g(x) = 2x - 1 \). Substitute \( \frac{5}{4} \) into the function:\[g\left(\frac{5}{4}\right) = 2\left(\frac{5}{4}\right) - 1 = \frac{10}{4} - 1 = \frac{10}{4} - \frac{4}{4} = \frac{6}{4} = \frac{3}{2}\]

Step 3: Conclusion

The value of the expression \( g\left[h\left(-\frac{1}{2}\right)\right] \) is \( \frac{3}{2} \).

Key Concepts

Function Evaluation

Evaluating functions is a fundamental skill in algebra where you substitute a number (or another expression) into a function and simplify the expression to find a result. For instance, in the exercise provided, you start by evaluating the function \( h(x) = x^2 + 1 \) at \( x = -\frac{1}{2} \). This means replacing every \( x \) with \( -\frac{1}{2} \) in the expression. So, we compute \( h(-\frac{1}{2}) \) which gives us:

• Substituting \( -\frac{1}{2} \) leads to \[ h(-\frac{1}{2}) = \left(-\frac{1}{2}\right)^2 + 1 \]
• Simplifying, we perform the operations to get \( \frac{1}{4} + 1 \).
• This results in \( \frac{5}{4} \).

Function evaluation allows us to understand how functions behave with different inputs. It's an indispensable part of problem-solving in algebra.

Algebraic Functions

Algebraic functions are expressions that combine variables and numbers through operations like addition, subtraction, multiplication, division, and taking roots. In our exercise, the functions \( f(x) = 4x \), \( g(x) = 2x - 1 \), and \( h(x) = x^2 + 1 \) are algebraic functions because they include these operations.

• f(x): This is a linear function since it consists of multiplying \( x \) by 4.
• g(x): This is another linear function, but it subtracts 1 after multiplying \( x \) by 2.
• h(x): This is a quadratic function because it involves squaring \( x \) and then adding 1.

Understanding these functions' structures is crucial: linear functions graph as straight lines, while quadratic functions result in parabolic curves. Recognizing the type of algebraic function helps you predict their behavior, aiding in effective problem-solving.

Problem Solving in Algebra

Algebra is all about solving problems by performing operations on expressions or finding unknown values. Problem-solving often involves multiple steps and a good understanding of algebraic principles. In this exercise, the problem is solved through function composition:

• First, we determine \( h(-\frac{1}{2}) \), which acts as the 'output' or solution of the first part and becomes the 'input' for the second function.
• We then evaluate \( g(x) \) using the result from \( h(x) \), showing how one function's output can directly affect another’s input, forming a cycle of operations.
• The final result, \( g(h(-\frac{1}{2})) \), combines these steps into a single solution: \( \frac{3}{2} \).

Breaking down complex problems into manageable steps simplifies the process, allowing you to systematically solve even the trickiest of algebraic problems. Mastering these techniques is crucial for success in algebra and mathematics in general.