Problem 78
Question
Express the given function \(h\) as a composition of two functions \(f\) and \(g\) so that \(h(x)=(f \circ g)(x)\). $$h(x)=\sqrt{5 x^{2}+3}$$
Step-by-Step Solution
Verified Answer
The function \(h(x)=\sqrt{5 x^{2}+3}\) can be expressed as a composition of two functions \(f\) and \(g\) so that \(h(x)=(f \circ g)(x)\), where \(f(x)=\sqrt{x}\) and \(g(x)=5x^{2}+3\).
1Step 1: Identify the Inner Function
Start by finding the 'inner function'. This is the operation being processed first according to the order of operations in algebra (Brackets, Orders, Division and Multiplication, Addition and Subtraction - BODMAS). Here, the inner function is the operation inside the square root, which is \(5 x^{2}+3\). So let's define our inner function \(g(x)=5x^{2}+3\).
2Step 2: Identify the Outer Function
Next, identify the 'outer function'. This is the operation being processed after the inner function. In this case, it's the square root operation. So let's define our outer function \(f(x)=\sqrt{x}\).
3Step 3: Compose the Functions
Now, to form our original function \(h\) as a composition of \(f\) and \(g\), we substitute \(g(x)\) into \(f(x)\) in place of \(x\) in the outer function. This would give us \(f(g(x)) = f(5x^{2}+3) = \sqrt{5x^{2}+3}\).
Key Concepts
Inner FunctionOuter FunctionOrder of Operations
Inner Function
In function composition, understanding the inner function is crucial because it sets the stage for what happens next. Think of it as the initial operation. For a function like \( h(x) = \sqrt{5x^2 + 3} \), we identify the inner function by focusing on what's inside the square root. This part is like the foundation or the preparatory step that everything else is built upon.
For this example, the expression inside the square root \( 5x^2 + 3 \) is the inner function, represented as \( g(x) = 5x^2 + 3 \). What makes the inner function special is that it's the first step in evaluating the entire composed function. It's like creating the batter before baking a cake.
To put it simply, the inner function is what you calculate first. Think about:
For this example, the expression inside the square root \( 5x^2 + 3 \) is the inner function, represented as \( g(x) = 5x^2 + 3 \). What makes the inner function special is that it's the first step in evaluating the entire composed function. It's like creating the batter before baking a cake.
To put it simply, the inner function is what you calculate first. Think about:
- What goes on inside any brackets or specific operations that happen first?
- How does the inside transform before applying any other operations?
Outer Function
After identifying the inner function, the next step is to determine the outer function. The outer function processes the result of the inner function. It's the broader context that frames how we interpret the result of our first operation, further transforming it to achieve the final output.
In our example, once we know \( g(x) = 5x^2 + 3 \), we need a function that takes this result and applies another operation. This is where the square root operation comes in, represented as the outer function \( f(x) = \sqrt{x} \).
The outer function can be seen as the final touch on your masterpiece. Just after preparing the ingredients (inner function), you bake it to perfection with the outer function to get the result you desire. You can think of the outer function as the wrapping paper that goes around a pre-packaged gift. The ingredients inside were prepared using the inner function, while the outer function wraps it all up neatly.
When dealing with the outer function, ask yourself:
In our example, once we know \( g(x) = 5x^2 + 3 \), we need a function that takes this result and applies another operation. This is where the square root operation comes in, represented as the outer function \( f(x) = \sqrt{x} \).
The outer function can be seen as the final touch on your masterpiece. Just after preparing the ingredients (inner function), you bake it to perfection with the outer function to get the result you desire. You can think of the outer function as the wrapping paper that goes around a pre-packaged gift. The ingredients inside were prepared using the inner function, while the outer function wraps it all up neatly.
When dealing with the outer function, ask yourself:
- How will the result from the inner function be further processed or altered?
- What is the operation or function applied to the initial result?
Order of Operations
The order of operations is a systematic way to solve expressions and equations, which guarantees consistency and correctness. It's commonly remembered by the acronym PEMDAS (Parentheses, Exponents, Multiplication and Division, Addition and Subtraction) or BODMAS (Brackets, Orders, Division and Multiplication, Addition and Subtraction). These help us decide which operations to perform first.
In function composition, understanding the order of operations is key to identifying inner and outer functions. Think of it as a roadmap: it shows you the sequence of steps you need to follow to reach your destination. In our example, \( h(x) = \sqrt{5x^2 + 3} \), the inner operations include squaring \( x \) and performing the addition. Once these are done, the square root (outer operation) is performed.
The order of operations ensures that calculations are predictable and clear. Here's a quick checklist to help when you're composing functions:
In function composition, understanding the order of operations is key to identifying inner and outer functions. Think of it as a roadmap: it shows you the sequence of steps you need to follow to reach your destination. In our example, \( h(x) = \sqrt{5x^2 + 3} \), the inner operations include squaring \( x \) and performing the addition. Once these are done, the square root (outer operation) is performed.
The order of operations ensures that calculations are predictable and clear. Here's a quick checklist to help when you're composing functions:
- First, tackle any operations inside parentheses or brackets.
- Next, resolve exponents or orders.
- Then, handle multiplication and division from left to right.
- Finally, take care of addition and subtraction from left to right.
Other exercises in this chapter
Problem 78
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