Problem 5 - College Algebra and Calculus: An Applied Approach

Question

Identify the inside function, \(u=g(x)\), and the outside function, \(y=f(u)\). $$ y=f(g(x)) \quad u=g(x) \quad y=f(u) $$ $$ y=\sqrt{5 x-2} $$

Step-by-Step Solution

Answer

The inside function \(u = g(x) = 5x - 2\) and the outside function \(y = f(u) = \sqrt{u}\)

Step 1: Identify the Whole Function

The entire function given is \( y=\sqrt{5 x-2} \). So, this is our f(g(x)).

Step 2: Identify the inside function

By inspecting the function, we notice that within the square root, the function used is \(5x - 2\). So this is our g(x), hence \( u = g(x) = 5x -2\).

Step 3: Identifying the outside function

The outside function is the function that is applied to \(u = g(x)\), which in this case is the square root function. So our \(f(u) = \sqrt{u} \).

Key Concepts

Composition of Functions

When dealing with complex mathematical functions, we often have functions composed of other functions. This is known as the "composition of functions".
In mathematical notation, it is denoted as \(f(g(x))\), where \(g(x)\) is the inside function, and \(f(u)\) is the outside function when \(u=g(x)\).
These compositions allow us to build more layered and intricate functions from simpler ones. Let's break it down:

• The inside function \(g(x)\): It is the function that is evaluated first. It forms the input for the outside function.
• The outside function \(f(u)\): This function acts on the result of the inside function, producing the final output.

For example, if our function is represented as \(y=\sqrt{5x-2}\), the inside function here is \(g(x)=5x-2\) and the outside is \(f(u)=\sqrt{u}\).Understanding this concept is crucial to applying differentiation techniques effectively, as later sections will show.

Square Root Function

The square root function is a fundamental concept in mathematics known for taking a non-negative number and returning another non-negative number whose square is the given number.
In terms of differentiation, it's crucial to recognize that the square root function can be represented as a power, to be specific: \( \sqrt{x} = x^{1/2} \).
This power form makes it easier to apply calculus techniques.

• The general format of square root functions is \(f(x) = \sqrt{x}\), but it can also appear as \(\sqrt{g(x)}\) when part of a composition.
• Always remember that you can rewrite square roots using powers, which is handy for deriving functions.

In the context of our example \(y=\sqrt{5x-2}\), the square root represents the outside function, applying the operation to the result from \(g(x)=5x-2\).
This understanding is pivotal when diving into differentiation techniques, particularly the chain rule.

Differentiation Techniques

Differentiation involves finding the rate at which a function changes. A key technique in differentiation is the chain rule, especially applicable in composite functions.
The chain rule asserts that the derivative of a composite function \(f(g(x))\) is the derivative of the outside function evaluated at the inside function, multiplied by the derivative of the inside function.In our example of the function \(y=\sqrt{5x-2}\), we use the chain rule:

• The outside function is \(\sqrt{u}\). Differentiating \(\sqrt{u} = u^{1/2}\) gives \(\frac{1}{2}u^{-1/2}\).
• The inside function, \(g(x)=5x-2\), has a derivative of \(g'(x) = 5\).
• Using the chain rule, the derivative \(y'\) would be \(\frac{1}{2}(5x-2)^{-1/2} \times 5\).

Breaking it down allows us to handle more complicated functions with ease. Always ensure you use appropriate differentiation strategies, adapting to the form of the functions at hand.
This will enable you to approach a wide variety of problems with confidence.