Problem 42 - Algebra 2

Question

If \(f(x)=4 x, g(x)=2 x-1,\) and \(h(x)=x^{2}+1,\) find each value. $$ [g \circ(f \circ h)](3) $$

Step-by-Step Solution

Answer

The value of \([g \circ(f \circ h)](3)\) is 79.

Step 1: Understand the Composition of Functions

The problem asks us to find \([g \circ(f \circ h)](3)\). This notation indicates that we first apply \(h(x)\), then apply \(f(x)\) to the result of \(h(x)\), and finally apply \(g(x)\) to the outcome of \(f(x)\). The function composition order is critical: start with \(h\), proceed to \(f\), and finish with \(g\).

Step 2: Calculate \(h(3)\)

The function \(h(x)\) is given by \(x^2 + 1\). To find \(h(3)\), substitute 3 into this equation: \(h(3) = 3^2 + 1 = 9 + 1 = 10\). This is the output of the function \(h\).

Step 3: Find \(f(h(3))\) or \(f(10)\)

Next, apply the result from the previous step to \(f(x)\). The function \(f(x)\) is defined as \(4x\). Calculate \(f(10) = 4 imes 10 = 40\).

Step 4: Calculate \(g(f(h(3)))\) or \(g(40)\)

Lastly, apply \(g(x)\) to the result from the previous step. The function \(g(x)\) is \(2x - 1\). Therefore, calculate \(g(40) = 2 imes 40 - 1 = 80 - 1 = 79\).

Step 5: Conclusion

After completing each composition step, we find that \([g \circ(f \circ h)](3) = 79\). This final value is the result of the entire function composition.

Key Concepts

Function Operations

Understanding function operations is fundamental when dealing with combinations of different functions in mathematics. At its core, function operation involves applying one function to the result of another. In this exercise, you'll encounter three types of operations: addition, subtraction, multiplication, and division of functions. However, the unique operation here is the composition of functions, which involves combining them in a sequential manner.

The idea is simple: you take the output of one function and use it as the input for another. This chaining process allows functions to interact and perform a series of mathematical transformations. The function order is crucial, just like following a recipe step by step. For example:

• Start with the function h, simplify it, and use its result.
• Pass the outcome through f to get a new result.
• Finish with g using the result from f as its input.

Composite Function Evaluation

When working on composite functions, the notation like \(g \circ (f \circ h)\) can initially look intimidating. However, understanding it can become straightforward with familiarity.

In the expression \(g \circ(f \circ h)\), we do the operations in the reverse of how they're written. Start by substituting your number into the innermost function, which is \(h(x)\). This first step simplifies to find the result of \(h(3)\). The output of this step becomes the input of the next function.

• The outcome of \(h(3)\) is used as the input for \(f\), resulting in \(f(h(3))\) or \(f(10)\).
• Finally, \(g\) uses this result to produce \(g(f(h(3)))\) or \(g(40)\).

Each step builds on results from the prior step, making composite function evaluation a neat transformation process.

Algebraic Expressions

Algebraic expressions are indispensable tools in function operations, allowing us to write out operations clearly and perform calculations effectively. In the problem at hand, each function is defined using an algebraic expression:

• Function f(x) is represented by \(4x\).
• Function g(x) is written as \(2x - 1\).
• Function h(x) is given by \(x^2 + 1\).

Understanding these expressions allows you to substitute specific values and compute the outputs methodically. In the composition process, recognizing these expressions helps maintain clarity so that each substitution and operation aligns with mathematical rules.

For instance, simplifying \(h(3)\) from \(h(x) = x^2 + 1\) gives us a solid start, showing how powerful and manageable algebraic expressions can be when navigating through function compositions.