Problem 25
Question
(a) Graph \(f(x)=x^{2}\) and \(g(x)=x^{3}\) for \(x \geq 0\), together, in one coordinate system. (b) Show algebraically that $$ x^{2} \geq x^{3} $$ for \(0 \leq x \leq 1\) (c) Show algebraically that $$ x^{2} \leq x^{3} $$ for \(x \geq 1\).
Step-by-Step Solution
Verified Answer
For \( x \geq 0 \), \( x^2 \geq x^3 \) for \( 0 \leq x \leq 1 \) and \( x^2 \leq x^3 \) for \( x \geq 1 \).
1Step 1: Understand the Functions
Consider the two functions: - \( f(x) = x^2 \) which is a parabola opening upwards. - \( g(x) = x^3 \) which is a cubic function with an inflection point at the origin. We will graph these functions from \( x = 0 \) onwards.
2Step 2: Sketch the Graphs
Graph \( f(x) = x^2 \) and \( g(x) = x^3 \) for \( x \geq 0 \). For positive \( x \), the parabola \( f(x) \) will lie above the cubic \( g(x) \) near the origin and will intersect at \( x = 1 \). Beyond that, \( g(x) \) grows faster than \( f(x) \).
3Step 3: Show \(x^2 \geq x^3\) for \(0 \leq x \leq 1\)
Consider the inequality: \[ x^2 \geq x^3 \] This can be rewritten as: \[ x^2 - x^3 \geq 0 \] Factor the expression: \[ x^2(1 - x) \geq 0 \] For \(0 \leq x \leq 1\), \( x^2 \geq 0 \) and \( 1 - x \geq 0 \), making the product non-negative.
4Step 4: Show \(x^2 \leq x^3\) for \(x \geq 1\)
Consider the inequality: \[ x^2 \leq x^3 \] Rewrite as: \[ x^2 - x^3 \leq 0 \] Factor the expression: \[ x^2(x - 1) \leq 0 \] For \(x \geq 1\), \( x^2 \geq 0 \) and \( x - 1 \geq 0 \), Thus the product is non-positive.
Key Concepts
Understanding FunctionsExploring InequalitiesGraphing FunctionsAlgebraic Proof
Understanding Functions
Functions play a crucial role in calculus and mathematics in general. They map inputs to outputs and are often represented as mathematical expressions. In this exercise, we explore two specific functions:
- Quadratic Function: The function \( f(x) = x^2 \) is quadratic. It is represented by a parabola opening upwards. This means as \( x \) increases, \( f(x) \) grows, initially at a slower rate and then more rapidly.
- Cubic Function: The function \( g(x) = x^3 \) is cubic. It shows a characteristic inflection point at the origin. This means its rate of growth changes and becomes more pronounced as \( x \) moves away from zero.
Exploring Inequalities
Inequalities are mathematical expressions that determine the relative size of two values. We use inequalities to understand where one function is larger than the other. In this exercise, we examine two specific inequalities:
- For \( 0 \leq x \leq 1 \): The inequality \( x^2 \geq x^3 \) implies that the quadratic function is greater than or equal to the cubic function in this domain. We start by manipulating the inequality into the form: \[ x^2 - x^3 \geq 0 \] By factoring, we get: \[ x^2(1 - x) \geq 0 \]
- For \( x \geq 1 \): Conversely, the inequality \( x^2 \leq x^3 \) indicates that beyond \( x = 1 \), the cubic function overtakes the quadratic. It's expressed as: \[ x^2(x - 1) \leq 0 \]
Graphing Functions
Graphing is a visual way of understanding the behavior of functions. It helps us see where functions intersect, which proves useful for analyzing inequalities. For this exercise, we are plotting the functions \( f(x) = x^2 \) and \( g(x) = x^3 \) for \( x \geq 0 \).
- Begin by marking key points, such as intersections, which for \( f(x) \) and \( g(x) \) occurs at \( x = 1 \).
- The quadratic parabola starts at the origin and opens upwards, whereas the cubic curve also starts at the origin but eventually grows faster as \( x \) increases.
Algebraic Proof
Algebraic proofs involve showing the truth of mathematical statements using algebraic manipulation. They are systematic and often involve steps such as expansion and factoring:
- For \( 0 \leq x \leq 1 \): We need to show \( x^2 \geq x^3 \). This transforms into \[ x^2 - x^3 = x^2(1 - x) \geq 0 \] Here, both \( x^2 \) and \( 1-x \) are non-negative within the given range, which makes their product non-negative.
- For \( x \geq 1 \): The proof involves \( x^2 \leq x^3 \). This transforms into \[ x^2(x - 1) \leq 0 \] Here, while \( x^2 \geq 0 \) is always true, \( x - 1 \ge 0 \) is true only for \( x \ge 1 \), altering the sign of the product.
Other exercises in this chapter
Problem 24
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