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Chapter 2: Problem 69

Sketch the graph of the given function \(f\) on the interval [-1.3,1.3]. $$ f(x)=x^{3}+1 $$

Short Answer

Expert verified
To sketch the graph of the function \(f(x) = x^{3} + 1\) on the interval [-1.3, 1.3], first find the intercepts: y-intercept is (0, 1) and x-intercept is (-1, 0). Since the function has an odd degree and a positive leading coefficient, it decreases to the left and increases to the right. Plot the intercepts, notice the "bump" in the interval [-1, 0], and connect the points with a smooth curve. This will give you the graph of the function in the given interval.

Step by step solution

01

Finding the Domain

The given function is a polynomial function and has no restrictions or discontinuities on its input. Therefore, the domain of the function is all real numbers.
02

Finding the y-intercept

To find the y-intercept, we set x = 0 and find the value of f(x). $$ f(0) = 0^3 + 1 $$ $$ f(0) = 1 $$ This means that the y-intercept of the function is (0, 1).
03

Finding the x-intercepts

To find the x-intercepts, we set f(x) = 0 and solve for x. $$ x^3 + 1 = 0 $$ $$ x^3 = -1 $$ $$ x = -1 $$ This means that the x-intercept of the function is (-1, 0).
04

Analyzing the End Behavior

Since the given function has an odd degree and a positive leading coefficient, the end behavior of the function will be as follows: 1. As \(x \to -\infty\), \(f(x) \to -\infty\) 2. As \(x \to \infty\), \(f(x) \to \infty\) This indicates that the graph will decrease to the left and increase to the right.
05

Sketching the Graph

Now that we have all the necessary information, we can sketch the graph of the function on the interval [-1.3, 1.3]. 1. Begin by plotting the intercepts, (-1, 0) and (0, 1). 2. Note that the graph decreases to the left and increases to the right, with a "bump" in the interval [-1, 0] 3. Connect the points with a smooth curve, making sure the graph passes through both intercepts and complies with the end behavior. With these steps, you should have sketched the graph of the given function, \(f(x) = x^{3} + 1\), on the interval [-1.3, 1.3].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Domain of Functions
The domain of a function is essentially the set of all possible input values, usually denoted as the variable "x." When discussing polynomial functions such as the one in this exercise, determining the domain is straightforward. Polynomials have no restrictions, meaning you can plug in any real number and the function will provide an output. Therefore, for the given function \(f(x) = x^3 + 1\), the domain is all real numbers. That means \((-\infty, \infty)\).

In simpler terms, this function will "accept" any number you choose to input. This characteristic is important in graphing because it ensures that the graph will cover every x-axis base point continuously without breaks.
Intercepts
Intercepts provide critical points that help outline the structure of a graph. There are two types of intercepts to consider:
  • Y-intercept: This occurs where the graph crosses the y-axis (where \(x = 0\)). For the function \(f(x) = x^3 + 1\), substituting \(x = 0\) gives us \(f(0) = 1\). Therefore, the y-intercept is at the point \((0, 1)\).
  • X-intercept: This occurs where the graph crosses the x-axis \((f(x) = 0)\). For our function, solving \(x^3 + 1 = 0\) results in \(x = -1\). Thus, the x-intercept is at \((-1, 0)\).
These intercepts are vital because they anchor the graph, showing where it emerges from one axis and potentially intersects with the other.
End Behavior
End behavior describes how the function behaves as the input values move towards positive or negative infinity. For polynomial functions, this behavior is dependent on both the degree and the sign of the leading coefficient, which influences the "direction" in which the graph heads.

In the function \(f(x) = x^3 + 1\), we have a polynomial of degree 3 with a positive leading coefficient (the coefficient of \(x^3\) is 1). This odd-degree polynomial ensures the graph will have opposite end behaviors:
  • As \(x \to -\infty\), \(f(x) \to -\infty\), meaning the graph descends to the left.
  • As \(x \to \infty\), \(f(x) \to \infty\), meaning the graph rises to the right.
Understanding this balance is key when predicting the graph's overall contour.
Polynomial Degree
The degree of a polynomial gives us a snapshot of the complexity and nature of its graph. The degree is simply the highest power of x in the polynomial. For \(f(x) = x^3 + 1\), the degree is 3 because the term \(x^3\) is the highest-powered term.

Here are some insights about a degree 3 polynomial:
  • It typically features an odd number of turning points and an odd-function mirror-symmetry about the origin in simpler forms.
  • With one bend or curve, it's what you expect from a cubic function, similar to an 'S' shape when exaggerated across wide ranges.
This information allows you to anticipate the general motion and style of the graph even before plotting points or calculating intercepts.When applied, the polynomial degree helps with sketching a visually accurate version of the function across its domain, completing analyses like those in this exercise.

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Most popular questions from this chapter

Suppose \(p(x)=a_{0}+a_{1} x+\cdots+a_{n} x^{n},\) where \(a_{0}, a_{1}, \ldots, a_{n}\) are integers. Suppose \(m\) is a nonzero integer that is a zero of \(p\). Show that \(a_{0} / m\) is an integer. [This result shows that to find integer zeros of a polynomial with integer coefficients, we need only look at divisors of its constant term.]

Find all real numbers \(x\) such that $$ x^{6}-8 x^{3}+15=0 $$.

Suppose $$r(x)=\frac{x+1}{x^{2}+3} \quad \text { and } \quad s(x)=\frac{x+2}{x^{2}+5}$$ What is the domain of \(r ?\)

Find all real numbers \(x\) such that $$ x^{4}-2 x^{2}-15=0 $$.

Write the indicated expression as a ratio of polynomials, assuming that $$ r(x)=\frac{3 x+4}{x^{2}+1}, \quad s(x)=\frac{x^{2}+2}{2 x-1}, \quad t(x)=\frac{5}{4 x^{3}+3} $$. $$ (s(x))^{2} $$
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