Problem 28 - Algebra 2

Question

Find \([g \circ h](x)\) and \([h \circ g](x)\) $$ \begin{array}{l}{g(x)=4 x} \\ {h(x)=2 x-1}\end{array} $$

Step-by-Step Solution

Answer

\([g \circ h](x) = 8x - 4\) and \([h \circ g](x) = 8x - 1\).

Step 1: Understand Composition of Functions

When you see \([g \circ h](x)\), it means you need to substitute the entire function \(h(x)\) into every \(x\) of the function \(g(x)\). Similarly, \([h \circ g](x)\) means substituting the function \(g(x)\) into every \(x\) of the function \(h(x)\).

Step 2: Substitute h(x) into g(x) for [g \circ h](x)

The function \(g(x)\) is defined as \(4x\). To find \([g \circ h](x)\), substitute \(h(x) = 2x - 1\) into \(g(x)\):\[\begin{align*}\ [g \circ h](x) &= g(h(x)) \&= g(2x - 1) \&= 4(2x - 1).\end{align*}\]

Step 3: Simplify for [g \circ h](x)

Simplify the expression by distributing 4: \[\begin{align*}\ 4(2x - 1) = 4 \times 2x - 4 \times 1 = 8x - 4.\end{align*}\] Hence, \([g \circ h](x) = 8x - 4\).

Step 4: Substitute g(x) into h(x) for [h \circ g](x)

Now let's find \([h \circ g](x)\) by substituting \(g(x) = 4x\) into \(h(x)\):\[\begin{align*}\ [h \circ g](x) &= h(g(x)) \&= h(4x) \&= 2(4x) - 1.\end{align*}\]

Step 5: Simplify for [h \circ g](x)

Simplify the expression by multiplying: \[\begin{align*}\ 2(4x) - 1 = 8x - 1.\end{align*}\] Thus, \([h \circ g](x) = 8x - 1\).

Key Concepts

Algebraic Functions

Algebraic functions are mathematical expressions that involve operations such as addition, subtraction, multiplication, division, and the application of powers or roots to variables. In the exercise you've seen, both functions \(g(x) = 4x\) and \(h(x) = 2x - 1\) are linear, which means their outputs change at a constant rate as their inputs change.

In deeper mathematical terms:

• \(g(x) = 4x\) is a simple linear function where for each unit increase in \(x\), \(g(x)\) increases by 4.
• \(h(x) = 2x - 1\) is also linear, where \(h(x)\) increases by 2 for each unit increase in \(x\), but it is shifted downwards by 1 on the y-axis due to the \(-1\).

Understanding these basic algebraic manipulations helps set the stage for exploring more complex relationships such as those found in composite functions.

Composite Functions

Composite functions involve taking two functions and combining them to make a new function. It's like stacking two processes together: one happens first, followed by the second. This is represented by the notation \([g \circ h](x)\) or \([h \circ g](x)\).

• For \([g \circ h](x)\), you first plug \(x\) into \(h(x)\) and then take the result to plug into \(g(x)\).
• For \([h \circ g](x)\), you switch the order. First, plug \(x\) into \(g(x)\) and then take that result to plug into \(h(x)\).

Let's see it in action:- To find \([g \circ h](x)\), substitute \(2x - 1\) from \(h(x)\) into \(g(x) = 4x\), giving us the expression \(4(2x - 1)\).- For \([h \circ g](x)\), substitute \(4x\) from \(g(x)\) into \(h(x) = 2x - 1\), resulting in \(2(4x) - 1\).

Remember, the order of operation is essential in composition, as changing the order changes the outcome.

Function Simplification

Simplifying functions is all about reducing complex expressions into their simplest form to ease interpretations and calculations. It involves performing basic algebraic manipulations such as distributing, combining like terms, and canceling out terms where possible.

In our example:

• For \([g \circ h](x) = 4(2x - 1)\), distribute the 4 to get \(4 \cdot 2x - 4 \cdot 1 = 8x - 4\).
• For \([h \circ g](x) = 2(4x) - 1\), distribute the 2 to get \(2 \cdot 4x - 1 = 8x - 1\).

This simplification helps in understanding the impact each variable or constant has on the equations, making the overall behavior of functions more evident. Reducing expressions to a simple form is a powerful tool in solving, analyzing, and graphing functions.