Problem 39 - Algebra and Trigonometry with Analytic Geometry

Question

If \(D(t)=\sqrt{400+t^{2}}\) and \(R(x)=20 x\), find \((D \circ R)(x)\)

Step-by-Step Solution

Answer

\((D \circ R)(x) = 20 \sqrt{1 + x^2}\)

Step 1: Understand Composition of Functions

The task requires finding the composition of two functions, given as \( (D \circ R)(x) \). This means we need to substitute \( R(x) \) into \( D(t) \). This is interpreted as \( D(R(x)) \).

Step 2: Identify and Substitute \( R(x) \) Into \( D(t) \)

First, identify the two functions:- \( D(t) = \sqrt{400 + t^2} \)- \( R(x) = 20x \)Now, substitute \( R(x) \) into \( D(t) \) to get \( D(R(x)) = D(20x) \).

Step 3: Simplify \( D(R(x)) \)

Substitute \( t = 20x \) into \( D(t) \):\[ D(20x) = \sqrt{400 + (20x)^2} \]Calculate \((20x)^2 = 400x^2\) and substitute:\[ D(20x) = \sqrt{400 + 400x^2} \].

Step 4: Simplify the Expression

Combine the terms under the square root:\[ \sqrt{400 + 400x^2} = \sqrt{400(1 + x^2)} \]Factor out 400 from the square root:\[ = \sqrt{400} \times \sqrt{1 + x^2} \].

Step 5: Simplify Further

Note that \( \sqrt{400} = 20 \). So, simplify the expression:\[ D(20x) = 20 \times \sqrt{1 + x^2} \].

Step 6: Conclusion

The composition \( (D \circ R)(x) = 20 \sqrt{1 + x^2} \). This expression gives the simplified form of the function composition.

Key Concepts

Square Root Function

The square root function is one of the fundamental functions in mathematics. When you see a function like \( f(x) = \sqrt{x} \), it tells you to look for a value that, when multiplied by itself, results in \( x \). In our original exercise, the square root function comes into play as part of \( D(t) = \sqrt{400 + t^2} \). This function suggests taking the square root of the expression inside it. Here's how to approach it:

• Identify what's inside the square root. In this case, it's \( 400 + t^2 \).
• Understand that the square root applies to this whole expression.
• When any variable expression, such as \( t \), is included, separate it and consider how its square impacts the overall value under the root.

Breaking down the square root function can make complex expressions simpler to handle, especially when performing operations like composition.

Composition of Functions

Composition of functions, often denoted as \((D \circ R)(x)\), is a process where you apply one function to the results of another. This is a crucial concept for building complex functions from simple ones.For the exercise at hand, we have to understand that:

• \( D(t) = \sqrt{400 + t^2} \) and \( R(x) = 20x \) are two separate functions.
• When composing \( D \) and \( R \), you need to substitute \( R(x) \) into \( D(t) \), effectively computing \( D(R(x)) \).

In more straightforward terms, replace ‘\( t \)’ in \( D(t) \) with the expression you have for \( R(x) \). Following these steps helps avoid mistakes and aligns your function composition process with your math goals.Understanding function composition allows you to explore more dynamic relationships between different variables and expressions in algebra.

Algebraic Simplification

Algebraic simplification is about making an expression as straightforward as possible. In the context of our exercise, this entails manipulating the expression \( \sqrt{400 + 400x^2} \) to a simpler form.Here's how you simplify:

• Recognize common factors, such as 400 in our exercise. It is part of both terms in \( 400 + 400x^2 \).
• Factor out to streamline the expression inside the square root: \( \sqrt{400(1 + x^2)} \).
• Apply properties of square roots that allow you to split the square root over a product: \( \sqrt{400} \times \sqrt{1 + x^2} \).
• Simplify \( \sqrt{400} = 20 \), resulting in the expression \( 20 \sqrt{1 + x^2} \).

A firm grasp on algebraic simplification facilitates evaluating expressions quickly and understanding the underlying structure. This skill is crucial for solving more intricate problems efficiently.