Problem 62
Question
Find \(f(x)\) at the indicated value of \(x\). $$f(x)=\sqrt[3]{x^{2}-x+6}, x=2$$
Step-by-Step Solution
Verified Answer
The value of \( f(2) \) is 2.
1Step 1: Write Down the Function and Value of x
We start with the function given in the problem: \( f(x) = \sqrt[3]{x^2 - x + 6} \). We are asked to find the value of this function at \( x = 2 \).
2Step 2: Substitute x into the Function
Replace \( x \) with 2 in the expression \( x^2 - x + 6 \). This gives: \( 2^2 - 2 + 6 \).
3Step 3: Simplify the Expression inside the Root
Calculate the expression inside the cube root: \( 2^2 = 4 \), so \( 4 - 2 + 6 = 8 \).
4Step 4: Calculate the Cube Root
Find the cube root of 8: \( \sqrt[3]{8} = 2 \) since \( 2^3 = 8 \).
5Step 5: Write the Solution
Thus, \( f(2) = 2 \).
Key Concepts
Cube RootSubstitutionSimplificationAlgebraic Functions
Cube Root
The cube root of a number is a value that, when multiplied by itself twice, gives the original number. In mathematical notation, the cube root of a number \(a\) is written as \(\sqrt[3]{a}\). For example, the cube root of 8 is 2, because \(2 \times 2 \times 2 = 8\). Understanding the cube root is crucial when working with algebraic expressions, as it allows us to simplify equations involving cubic powers.
In practical terms, finding the cube root is about determining which number, when cubed, achieves a given result. The cube root is also visually recognizable by the symbol \(\sqrt[3]{...}\), which indicates it's a root of the third degree. This contrasts from the square root \(\sqrt{...}\), a second degree root, which you may be more familiar with.
In our exercise, you've seen how finding the cube root helped simplify a seemingly complex expression to a straightforward number. Calculating cube roots involves both practice and understanding of the basic powers of numbers.
In practical terms, finding the cube root is about determining which number, when cubed, achieves a given result. The cube root is also visually recognizable by the symbol \(\sqrt[3]{...}\), which indicates it's a root of the third degree. This contrasts from the square root \(\sqrt{...}\), a second degree root, which you may be more familiar with.
In our exercise, you've seen how finding the cube root helped simplify a seemingly complex expression to a straightforward number. Calculating cube roots involves both practice and understanding of the basic powers of numbers.
Substitution
Substitution in mathematics involves replacing a variable in an expression with a given value, which lets us evaluate the expression at a specific point. In our exercise, we were given a function, \(f(x) = \sqrt[3]{x^2 - x + 6}\), and a point \(x = 2\). The task was to find \(f(2)\).
The substitution was simple; we replaced every instance of \(x\) in the expression \(x^2 - x + 6\) with 2. Here's how it looked: the expression \(x^2 - x + 6\) became \(2^2 - 2 + 6\). Substitution like this is a foundational skill in algebra, as it allows us to find specific results and understand how variables affect an equation.
It's helpful to think of substitution as a simple plug-and-play action. You're placing a number directly into the spots where variables appear, thus allowing further evaluation.
The substitution was simple; we replaced every instance of \(x\) in the expression \(x^2 - x + 6\) with 2. Here's how it looked: the expression \(x^2 - x + 6\) became \(2^2 - 2 + 6\). Substitution like this is a foundational skill in algebra, as it allows us to find specific results and understand how variables affect an equation.
It's helpful to think of substitution as a simple plug-and-play action. You're placing a number directly into the spots where variables appear, thus allowing further evaluation.
Simplification
Simplification is the process of reducing an expression to its simplest form. This typically involves performing arithmetic operations, combining like terms, or making use of arithmetic properties. Within the exercise, after substitution, the expression \(2^2 - 2 + 6\) needed to be simplified.
When we simplified, we calculated \(2^2\) which equals 4, resulting in the expression \(4 - 2 + 6\). By following straightforward addition and subtraction, the values were combined to reach 8. Simplification is key in mathematics because it clarifies the expression, leading to easier interpretation and further evaluation.
When we simplified, we calculated \(2^2\) which equals 4, resulting in the expression \(4 - 2 + 6\). By following straightforward addition and subtraction, the values were combined to reach 8. Simplification is key in mathematics because it clarifies the expression, leading to easier interpretation and further evaluation.
- Breaking issues into manageable parts simplifies complexity.
- Combining like terms reduces unnecessary components.
- Evaluating expressions at each step ensures that no errors are made.
Algebraic Functions
Algebraic functions include expressions composed of variables and constants with operations like addition, subtraction, multiplication, division, and roots. In our example, the function \(f(x) = \sqrt[3]{x^2 - x + 6}\) mixes these elements through squaring variables and employing the cube root.
Algebraic functions are central to algebra because they represent mathematical relationships and can describe real-world phenomena. When evaluating algebraic functions, understanding the role of each term is essential. Our function combined a quadratic polynomial \(x^2 - x + 6\) with a cube root operation, illustrating how various algebraic operations intertwine.
Approaching algebraic functions:
Algebraic functions are central to algebra because they represent mathematical relationships and can describe real-world phenomena. When evaluating algebraic functions, understanding the role of each term is essential. Our function combined a quadratic polynomial \(x^2 - x + 6\) with a cube root operation, illustrating how various algebraic operations intertwine.
Approaching algebraic functions:
- Identify the type of equation or expression.
- Break down the expression into manageable components.
- Perform operations step by step like substitution and simplification to solve.
Other exercises in this chapter
Problem 61
Find a decimal approximation of each root or power. Round answers to the nearest thousandth. $$19^{1 / 2}$$
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Classify each equation as a contradiction, an identity, or a conditional equation. Give the solution set. Use a graph or table to support your answer. $$\frac{x
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Solve each formula for the specified variable.} \(V=L W H\) for \(L \quad\) (Volume of a box)
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Find a decimal approximation of each root or power. Round answers to the nearest thousandth. $$29^{1 / 3}$$
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