Problem 12 - Algebra A Combined Function

Question

If \(f(x)=x^{2}-6 x+2, g(x)=-2 x\), and \(h(x)=\sqrt{x}\), find each composition. $$ (f \circ h)(1) $$

Step-by-Step Solution

Answer

(f \circ h)(1) = -3.

Step 1: Understand the Composition

The composition \((f \circ h)(x)\) means you need to take \(h(x)\) and substitute it into \(f(x)\). So, it will be \(f(h(x))\).

Step 2: Calculate h(1)

The function \(h(x)\) is defined as \(\sqrt{x}\). Substitute \(x = 1\) into \(h(x)\) to get \(h(1) = \sqrt{1} = 1\).

Step 3: Substitute h(1) into f(x)

Now that we know \(h(1) = 1\), substitute 1 into \(f(x)\). The function \(f(x) = x^2 - 6x + 2\). Substitute to get \(f(1) = (1)^2 - 6(1) + 2\).

Step 4: Calculate f(1)

Compute the expression \(1 - 6 + 2\). Simplifying this yields: \(1 - 6 = -5\); then, \(-5 + 2 = -3\). Therefore, \(f(1) = -3\).

Key Concepts

Function Notation

Function notation is a way of representing functions in a concise and consistent manner. It typically uses a format like \( f(x) \), where \( f \) represents the function and \( x \) is the input. Knowing how to read function notation is essential to solving composition problems.
This notation helps identify the relationship between variables and the operations that must be performed. In the given exercise, we have functions represented as \( f(x) = x^2 - 6x + 2 \), \( g(x) = -2x \), and \( h(x) = \sqrt{x} \). Each function has a different mathematical rule applied to the input \( x \).

• \( f(x) \) squares \( x \) and modifies it with a linear expression \(-6x + 2\).
• \( g(x) \) multiplies \( x \) by \(-2\).
• \( h(x) \) takes the square root of \( x \).

Function notation not only makes the expression of these rules clear but also makes operations like composition straightforward.

Substitution Method

The substitution method is a key strategy for working with function compositions. It involves substituting the output of one function into another.
In the problem, we're asked to find \((f \circ h)(1)\), which is a composition requiring us to substitute \( h(1) \) into \( f(x) \).
This approach is systematic and can be broken down into simple steps:

• First, determine \( h(1) \), which involves substituting the input value \( 1 \) into the function \( h(x) = \sqrt{x} \). This gives us \( h(1) = \sqrt{1} = 1 \).
• Next, substitute this result into \( f(x) \) to get \( f(1) \). Plug in \( x = 1 \) into \( f(x) = x^2 - 6x + 2 \).
• Simplify to find the final solution.

Substitution allows us to iterate step by step through functions, ensuring each function's operation is correctly applied.

Square Root Function

The square root function, denoted as \( h(x) = \sqrt{x} \), is a fundamental function in mathematics, impacting various applications from geometry to calculus.
Understanding the square root function involves recognizing that it finds a value which, when multiplied by itself, results in \( x \).

• The input \( x \) must be non-negative because square roots of negative numbers are not defined in the real number system.
• For \( x = 1 \), the function returns \( h(1) = \sqrt{1} = 1 \). This is straightforward since 1 is its own square root.

In our exercise, the square root function sets the groundwork for finding \( (f \circ h)(1) \) by initially transforming \( x = 1 \) into \( 1 \), which is then used in the substitution method for further computations in \( f(x) \). By understanding these basic properties, tackling compositions that involve \( \sqrt{x} \) becomes much more approachable.