Problem 20
Question
Given \(f(x)=x^{3}-1, \quad x>1\) \(=x-1, \quad x \leq 1\), find f(1)
Step-by-Step Solution
Verified Answer
Based on the given piecewise function, the relevant function for x=1 is f(x) = x-1. By substituting x=1 into the equation, we find that f(1) = 1 - 1 = 0. So, \( f(1) = 0 \).
1Step 1: Identify the relevant function for x=1
Since we want to find the value of f(1), we need to determine which part of the piecewise function is applicable for x=1. The condition for x=1 is given as x ≤ 1. According to the given function, when x≤1, f(x) = x-1.
2Step 2: Substitute x=1 into the relevant function
Now that we know the relevant function is f(x) = x-1, we can find the value of f(1) by substituting x=1 into this equation.
f(1) = 1 - 1
3Step 3: Calculate the value of f(1)
After substituting x=1 into the equation f(x) = x-1, we have:
f(1) = 1 - 1 = 0
Therefore, f(1) = 0.
Key Concepts
Function EvaluationConditional FunctionsAlgebraic Expressions
Function Evaluation
Understanding how to evaluate a function is a critical skill in mathematics, as functions are the backbone of many algebraic concepts. A function is a special kind of relationship where each input is associated with exactly one output. When we're asked to find the value of a function for a particular input, we're essentially plugging that input into the function's rule to find the output.
In the example, the function value for f(1) requires identifying which part of the piecewise function applies when x equals 1. This is a fundamental process in function evaluation: determining the correct rule to use based on the input value. As seen in the exercise, the function for x≤1 states that for these inputs, the output will be calculated using the rule f(x) = x - 1. So, by substituting the input 1 into the function, we effectively evaluate f(1) to get 0.
In the example, the function value for f(1) requires identifying which part of the piecewise function applies when x equals 1. This is a fundamental process in function evaluation: determining the correct rule to use based on the input value. As seen in the exercise, the function for x≤1 states that for these inputs, the output will be calculated using the rule f(x) = x - 1. So, by substituting the input 1 into the function, we effectively evaluate f(1) to get 0.
Conditional Functions
Piecewise functions, or conditional functions, are defined by multiple sub-functions, each with its own domain. These conditions determine which sub-function to use for a given input. The beauty of piecewise functions is that they can model complex, real-world scenarios that a single function rule could not capture.
In our example, the piecewise function consists of two different algebraic expressions assigned to different ranges of the variable x: one for x>1 and another for x≤1. When working with such functions, the first step is always to consider the conditions specified to determine which part of the function to employ. It's like following a map where different paths are taken based on where you stand. The correct evaluation of f(1) relies on recognizing that the condition x≤1 applies, thus leading to the correct expression and the final answer.
In our example, the piecewise function consists of two different algebraic expressions assigned to different ranges of the variable x: one for x>1 and another for x≤1. When working with such functions, the first step is always to consider the conditions specified to determine which part of the function to employ. It's like following a map where different paths are taken based on where you stand. The correct evaluation of f(1) relies on recognizing that the condition x≤1 applies, thus leading to the correct expression and the final answer.
Algebraic Expressions
At the heart of many mathematical problems, we find algebraic expressions. These are combinations of numbers, variables, and operations that represent a particular quantity. Mastering algebraic expressions involves understanding how to simplify them, evaluate them, and manipulate them according to algebraic principles.
In our step-by-step solution, we encounter two algebraic expressions: one is a cubic function, and the other is a linear function, each describing part of the piecewise function. The expression for x≤1 is linear and deceptively simple: f(x) = x - 1. To evaluate this expression for x=1, we replace x with 1 and perform the subtraction, a fundamental algebraic operation. This simplicity belies the importance of the process, however, as incorrect manipulation of algebraic expressions is a common source of error in mathematics. It's crucial to follow algebraic rules precisely for accurate problem-solving.
In our step-by-step solution, we encounter two algebraic expressions: one is a cubic function, and the other is a linear function, each describing part of the piecewise function. The expression for x≤1 is linear and deceptively simple: f(x) = x - 1. To evaluate this expression for x=1, we replace x with 1 and perform the subtraction, a fundamental algebraic operation. This simplicity belies the importance of the process, however, as incorrect manipulation of algebraic expressions is a common source of error in mathematics. It's crucial to follow algebraic rules precisely for accurate problem-solving.
Other exercises in this chapter
Problem 18
$$ \begin{aligned} &\text { Given }\\\ &f(x)=x^{2}, \quad x \geq 0\\\ &=0, \quad x
View solution Problem 19
$$ \begin{aligned} &\text { Given }\\\ &f(x)=x^{3}, \quad x \geq 0\\\ &=x^{2}, \quad x
View solution Problem 21
$$ \begin{aligned} &\text { Given }\\\ &\begin{aligned} f(x) &=e^{x}, \quad x \geq 0 \\ &=x+1, \quad x
View solution Problem 22
Given \(f(x)=\frac{x}{\sqrt{x^{2}}}, \quad x \neq 0\) \(=0, \quad x=0\), find \(f^{\prime}(0)\).
View solution