Problem 20 - Calculus Single Variable

Question

Let \(f(x)=\sqrt{2 x+5},\) and \(g(x)=x^{-1 / 3} .\) In Exercises \(19-22\) calculate the given expression. \((g \circ f)(2)\)

Step-by-Step Solution

Answer

\((g \circ f)(2) = \frac{1}{3^{1/3}}\).

Step 1: Understand the Composition of Functions

The expression \((g \circ f)(2)\) means that we need to find \(g(f(2))\). This requires us to first evaluate \(f(2)\) and then plug that result into the function \(g\).

Step 2: Evaluate f(2)

The function \(f(x) = \sqrt{2x+5}\). Substitute \(x = 2\) into the function: \[f(2) = \sqrt{2(2) + 5} = \sqrt{4 + 5} = \sqrt{9} = 3.\]

Step 3: Evaluate g(f(2))

With \(f(2) = 3\), we now need to find \(g(3)\) using the function \(g(x) = x^{-1/3}\): \[g(3) = 3^{-1/3}.\]

Step 4: Simplify g(3)

To simplify \(3^{-1/3}\), recall that \(a^{-b} = \frac{1}{a^{b}}\). Thus, \[g(3) = \frac{1}{3^{1/3}}.\]

Step 5: Final Expression

The result of the entire expression \((g \circ f)(2)\) is now simplified to \[\boxed{\frac{1}{3^{1/3}}}.\]

Key Concepts

Inverse Functions

In mathematics, an inverse function essentially reverses the effect of the original function. If you have a function that turns an input into an output, its inverse would take that output and bring you back to the original input.
A function is denoted as \(f(x)\), and its inverse is usually written as \(f^{-1}(x)\). For a function to have an inverse, each input must correspond to exactly one output, and each output must have originated from exactly one input. This is known as being bijective or one-to-one.

• An example is the function \(f(x) = y\). If \(f^{-1}(x) = y\), then swapping input and output returns you to the start.
• Not all functions have inverses. To find if a function has an inverse, it must pass the Horizontal Line Test, which tests for one-to-one nature.

Understanding the concept of inverse functions is fundamental because they provide a method to "undo" function operations in algebra, making them crucial tools in problem solving and calculus.

Function Evaluation

Function evaluation involves substituting specific values into a function to find the output. Essentially, you are "testing" the function with a particular input.
For example, consider the function \(f(x) = \sqrt{2x + 5}\). To evaluate \(f(2)\), you substitute \(2\) into \(x\), giving you \[f(2) = \sqrt{2(2) + 5} = \sqrt{9} = 3.\]

• The key steps in function evaluation are identifying the function form, substituting the given value, and simplifying.
• It is important to follow the order of operations (PEMDAS/BODMAS) for accurate results.
• Function evaluation is not only about numbers; it’s about understanding how inputs are transformed, which is crucial for deeper comprehension of mathematical models.

By consistently practicing the evaluation of functions, students become proficient at recognizing patterns and relationships in mathematics.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging expressions to simplify or solve them. It's like moving pieces of a puzzle to see the bigger picture.
This process includes actions such as distributing, factoring, combining like terms, and working with exponents and radicals. It involves using known algebraic identities and operations wisely.

• For instance, to simplify \(g(3) = 3^{-1/3}\), you apply the identity \(a^{-b} = \frac{1}{a^{b}}\), concluding \[g(3) = \frac{1}{3^{1/3}}.\]
• Manipulation allows one to handle complex expressions by breaking them down into simpler parts, which can be solved step by step.
• Proficiency in algebraic manipulation fuels success in higher-level math, such as calculus, where solving intricate equations is common.

Algebraic manipulation is a powerful tool for analyzing and interpreting mathematical situations, making it essential for students who wish to succeed in mathematics.