Problem 45 - Algebra and Trigonometry

Question

Find \(f \circ g \circ h\) $$ f(x)=x-1, \quad g(x)=\sqrt{x}, \quad h(x)=x-1 $$

Step-by-Step Solution

Answer

The composed function is \( f \circ g \circ h(x) = \sqrt{x - 1} - 1 \).

Step 1: Understand Composition Notation

The notation \(f \circ g \circ h\) means composing the functions. You first apply the innermost function \(h(x)\), then the result goes as input to the function \(g\), and finally the output of \(g\) becomes the input to \(f\).

Step 2: Evaluate h(x)

Begin by evaluating the innermost function in the composition, which is \( h(x) \). Given \( h(x) = x - 1 \), placeholder value for \( x \) yields \( h(x) = x - 1 \).

Step 3: Apply g to h(x)

Now take the result of the previous step and apply function \( g \). We know that \( g(x) = \sqrt{x} \), so substituting \( h(x) = x - 1 \) into \( g \) gives \( g(h(x)) = \sqrt{x - 1} \).

Step 4: Apply f to g(h(x))

Finally, use \( f(x) = x - 1 \) and substitute \( g(h(x)) = \sqrt{x - 1} \) into \( f \). Thus, \( f(g(h(x))) = \sqrt{x - 1} - 1 \). This completes the function composition \( f \circ g \circ h \).

Key Concepts

Composite Functions

Composite functions are formed when one function is applied to the result of another function. It can be visually understood using the notation \( f \circ g \circ h \). This may seem complex at first, but it breaks down into simpler parts by following a specific order.

• Start with the innermost function in the composition. In this exercise, it is \( h(x) \).
• Once you have the output of \( h(x) \), use it as the input for the next function, \( g \).
• Finally, take the result from \( g(h(x)) \) and use it as the input for \( f \).

The process of function composition is like a sequence of steps where each function builds on the result of the previous one. This ordered application allows us to effectively "compose" a new function from existing ones.

Algebra

Algebra is a branch of mathematics that deals with symbols and the rules for manipulating those symbols. In the context of composite functions, algebra helps us navigate through the process by using operations like substitution.When dealing with the exercise at hand, algebraic manipulation plays a crucial part.

• We see the function \( h(x) = x - 1 \), which is a simple linear function. By subtracting 1 from \( x \), it serves as the first step in our composition.
• The next function, \( g(x) = \sqrt{x} \), requires us to substitute \( h(x) \) as its input. Thus, we replace \( x \) with \( x - 1 \) inside the square root, transforming it into \( \sqrt{x - 1} \).
• Finally, the output of \( g(x) \) is fed into the last function, \( f \). With \( f(x) = x - 1 \), we substitute once more to derive the final expression \( \sqrt{x - 1} - 1 \).

Algebra thus becomes the language and framework for systematically expressing these relationships and transformations between functions, enabling us to solve and simplify composite functions.

Step-by-Step Solutions

Step-by-step solutions are pivotal in breaking down complex problems into manageable parts. They guide students in understanding how to approach and solve mathematical tasks, like composing functions, in a methodical way.In our exercise, this method helps by:

• Clarifying the order of operations. Always begin with the innermost function \( h(x) \), then proceed outward.
• Examining changes and substitutions at each stage, making it easier to follow and ensuring no errors are made.
• Simplifying the task which builds understanding from simpler problems to more complex solutions, reinforcing confidence in using composite functions.

A step-by-step approach ensures that details are not overlooked. It helps students identify where they might make mistakes and how to correct them, which fosters a deeper comprehension of mathematical procedures beyond rote memorization.