Problem 21 - 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. \(f^{2}(11) \cdot g^{3}(54)\)

Step-by-Step Solution

Answer

The answer is \(\frac{1}{2}\).

Step 1: Calculate \(f(11)\)

First, we need to find \(f(11)\). Plug \(x = 11\) into the function \(f(x) = \sqrt{2x + 5}\). Calculate: \[f(11) = \sqrt{2(11) + 5} = \sqrt{22 + 5} = \sqrt{27}.\] Thus, \(f(11) = \sqrt{27}.\)

Step 2: Calculate \(f^2(11)\)

Now, calculate \(f^2(11)\) which is the square of \(f(11)\): \[f^2(11) = (\sqrt{27})^2 = 27.\]

Step 3: Calculate \(g(54)\)

Next, substitute \(x = 54\) into the function \(g(x) = x^{-1/3}\): Calculate: \[g(54) = 54^{-1/3} = \frac{1}{\sqrt[3]{54}}.\]

Step 4: Calculate \(g^3(54)\)

We need to find \(g^3(54)\), which is the cube of \(g(54)\):\[g^3(54) = \left(\frac{1}{\sqrt[3]{54}}\right)^3 = \frac{1}{54}.\]

Step 5: Multiply the Results

Finally, we will multiply \(f^2(11)\) and \(g^3(54)\) together:\[f^2(11) \cdot g^3(54) = 27 \cdot \frac{1}{54} = \frac{27}{54} = \frac{1}{2}.\]

Key Concepts

Square Root Function

The square root function is a fundamental mathematical function represented as \( f(x) = \sqrt{x} \). It's crucial to understand that finding the square root of a number is the opposite of squaring that number. The square root of a number \( x \) is a value that, when multiplied by itself, gives \( x \). For example, if \( f(x) = \sqrt{2x + 5} \), and you input \( x = 11 \), you calculate:

• Substitute \( x \) into the function: \( \sqrt{2(11) + 5} \) which simplifies to \( \sqrt{27} \).
• The square root symbol, \( \sqrt{} \), indicates you are looking for a number that, when squared, equals the expression inside the root.

Understanding this concept helps you evaluate expressions involving square roots with ease and confidence.

Cube Root Function

The cube root function involves finding a number \( x \), such that when multiplied by itself three times, results in a certain value \( y \). This is represented as \( g(x) = x^{-1/3} \) or \( \sqrt[3]{x} \). The cube root is different from the square root, as it requires three factors instead of two. For example:

• For \( g(x) = x^{-1/3} \), substituting \( x = 54 \) gives you \( g(54) = 54^{-1/3} = \frac{1}{\sqrt[3]{54}} \).
• The cube root is particularly useful in solving problems where quantities are cubed or need to be divided into three equal parts.

It's essential in understanding functions involving division by three-fold factors and helps simplify complex problems involving cubes.

Composition of Functions

Composing functions involves creating a new function by applying one function to the results of another. Understanding this concept is essential when dealing with expressions formed from multiple functions. For example, if you have \( f(x) \) and \( g(x) \), composing these functions could mean finding \( (f \circ g)(x) \), which translates to \( f(g(x)) \).

• Composition requires you to first evaluate \( g(x) \), then use the result as the input for \( f(x) \).
• This concept builds a bridge between multiple layers of function applications, allowing for deeper problem-solving approaches.

Recognizing when and how to use function composition can greatly simplify complex problems and help in systematically breaking down challenging equations.

Multiplication of Functions

Multiplying functions often comes into play in advanced algebra calculations. This involves taking the product of two or more functions, typically denoted as \( (f \cdot g)(x) = f(x) \cdot g(x) \). For instance, when finding \( f^2(x) \cdot g^3(x) \):

• First, you evaluate \( f^2(x) \), which means squaring the result of \( f(x) \).
• Then, find \( g^3(x) \), the cube of \( g(x) \).
• The final step is multiplying these results together, in this case resulting in \( f^2(11) \cdot g^3(54) \).

Multiplying functions is helpful for combining different mathematical relationships and deriving new expressions from them in intricate algebra problems. It requires precision and understanding of each function's individual role.