Problem 19 - 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. \((f \circ g)(1 / 8)\)

Step-by-Step Solution

Answer

The value of \((f \circ g)(1/8)\) is 3.

Step 1: Understand Composition of Functions

The composition of two functions, \((f \circ g)(x)\), means that you first apply the function \(g(x)\) and then apply the function \(f(x)\) to the result of \(g(x)\). So, for \((f \circ g)(1/8)\), first find \(g(1/8)\) and then apply \(f\) to that result.

Step 2: Evaluate g(1/8)

The function \(g(x)\) is given as \(x^{-1/3}\). Substitute \(x = 1/8\) into \(g(x)\) to get:\[g(1/8) = (1/8)^{-1/3}\]Since \(x^{-1/3}\) is equivalent to \(1/(x^{1/3})\), simplify to get:\[g(1/8) = 1/(1/8)^{1/3} = 1/(1/2) = 2\]

Step 3: Calculate f(g(1/8))

Now that we know \(g(1/8) = 2\), apply the function \(f(x)\) to this result. The function \(f(x)\) is given by \(\sqrt{2x + 5}\). Substitute \(x = 2\) into \(f(x)\):\[f(2) = \sqrt{2(2) + 5} = \sqrt{4 + 5} = \sqrt{9} = 3\]

Step 4: Final Result

The expression \((f \circ g)(1/8)\) evaluates to \(3\). Therefore, the final answer is \(3\).

Key Concepts

Inverse Functions

Inverse functions are a fundamental concept in mathematics. They allow us to reverse the effects of a function, providing an output that corresponds to a given input.
In simple terms, if you have a function that takes an input and gives you an output, an inverse function will take that output and provide you back the original input.

• Consider a function that maps 2 to 4. Its inverse will map 4 back to 2. Mathematically, if "f" is our function, then its inverse, denoted as \( f^{-1} \), will reverse this process.
• A function must be bijective (both injective and surjective) to have an inverse. This means every element of the output is paired with an element of the input, and no elements are left unused or mapped twice.

Understanding inverse functions is crucial, especially when working with function compositions. They allow us to "undo" steps, giving new insights into the nature of the problems at hand.
When dealing with specific types such as inverse trigonometric or inverse exponential functions, knowing their properties and graphs makes solving problems, like those involving composition, much more intuitive.

Square Root Function

The square root function is a classic example of a mathematical function that provides the principal square root of a non-negative number. Each input to this function delivers only the non-negative, or principal, square root as its output.

• This function is symbolically represented as \( f(x) = \sqrt{x} \).
• The range of the square root function is non-negative real numbers, while its domain consists of non-negative numbers as well.

The square root function plays a critical role in many real-world scenarios. It's used in physics, engineering, and statistics to solve for variables that must stay non-negative due to practical constraints.
When combined with other functions in composition, it requires careful manipulation to maintain the integrity of the expressions involved. In our exercise, applying the square root function was the final step after transforming the initial input via the cubic root function. This demonstrates practicality in multi-step processes.

Cubic Root Function

The cubic root function, often denoted as \( g(x) = x^{1/3} \), is used to determine a value which, when cubed, returns the original number. Unlike the square root function, it can accept negative numbers and will yield both positive and negative roots.

• It is represented as: \( g(x) = x^{1/3} \), where \( x \) can be any real number.
• The domain and range of the cubic root function are both the set of all real numbers. This differs from the square root function, broadening the instances where it's applicable.

The versatility of the cubic root function allows for its use in more diverse mathematical and real-world contexts. In our exercise, calculating \( g(1/8) \) was essential to continue with function composition. Here, the cubic root function was utilized before applying the square root function, showcasing the layered structure of function compositions. Understanding how cubic roots work builds a foundation for handling complex numbers and polynomials efficiently.