/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 57 Graph the solution set of each s... [FREE SOLUTION] | 91Ó°ÊÓ

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Graph the solution set of each system of inequalities. $$\begin{aligned}&y>x^{3}+1\\\&y \geq-1\end{aligned}$$

Short Answer

Expert verified
The solution set is the region above the curve \(y = x^3 + 1\) and above the line \(y = -1\).

Step by step solution

01

Graph the Boundary of the First Inequality

Consider the equation of the boundary for the first inequality: \(y = x^3 + 1\). Graph this curve on a coordinate plane. This is a cubic function shifted 1 unit upward.
02

Determine the Region for the First Inequality

Since the first inequality is \(y > x^3 + 1\), shade the region above the curve \(y = x^3 + 1\). Use a dashed line for the boundary to indicate that points on the line are not included in the solution set.
03

Graph the Boundary of the Second Inequality

Next, consider the equation of the boundary for the second inequality: \(y = -1\). Graph this horizontal line on the coordinate plane.
04

Determine the Region for the Second Inequality

Since the second inequality is \(y \geq -1\), shade the region above and including the line \(y = -1\). Use a solid line for this boundary to indicate that points on the line are included in the solution set.
05

Find the Intersection of the Shaded Regions

The solution set for the system of inequalities is the region where the shaded areas from Step 2 and Step 4 overlap. This region will be above both \(y > x^3 + 1\) and \(y \geq -1\).

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Key Concepts

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

graphing inequalities
Graphing inequalities might seem tricky at first, but it’s all about understanding a few foundational concepts. When you graph an inequality like \( y > x^3 +1\) you start by graphing its boundary equation as if it were an equality, in this case, \( y = x^3 + 1\) . This helps us visualize the 'edge' of where the inequality starts.
Once you have the boundary, shading becomes key. Since our initial inequality is \( y > x^3 + 1\), we shade the area above the curve. Use a dashed line for the boundary curve to show that points on the curve are not part of the solution. For \( y \geq -1\), the procedure is similar, but we use a solid line as points on the line \( y = -1\) are included. Graphical representation aids in visual comprehension and making errors in arithmetic less likely.
cubic functions
Cubic functions, such as \( y = x^3 + 1\), are polynomial functions of degree three. They can exhibit a variety of behaviors that linear or quadratic functions can't, such as having an S-shaped curve. The cubic term dominates the function's behavior for large values of x, leading to the characteristic steep rise and fall.
When graphing cubic functions, it's helpful to identify key points. For instance, with \( y = x^3 + 1\), the graph shifts the basic cubic curve one unit up. Knowing how shifts affect the curve’s placement on the coordinate plane is crucial for accurately sketching the graph.
Cubic graphs tend to have a single inflection point where the curve changes concavity, allowing us to better predict the shape of the inequality regions.
intersection of regions
The intersection of regions approach comes into play when combining multiple inequalities. Once each inequality is graphed and its respective region shaded, the challenge is to find where these regions overlap. This overlapping part of the graph is where all inequalities are satisfied at the same time.
For our system, \( y > x^3 + 1 \) and \( y \geq -1 \), we first shade above the cubic curve and then above and on the horizontal line. The intersection will be the shaded part common to both, which effectively delineates our solution set.
Understanding intersections can be particularly useful in real-world applications like feasible region calculations in optimization problems or in defining areas of possible solutions in complex constraints.
coordinate plane
The coordinate plane forms the backdrop on which all graphing occurs. It consists of an x-axis (horizontal) and a y-axis (vertical) that intersect at the origin (0,0). This plane allows us to plot points, lines, curves, and shaded regions that represent various equations and inequalities.
When dealing with the coordinate plane, always start by drawing each axis and clearly marking the scale. For inequalities, understanding the plane’s quadrants (divisions created by the axes) helps in organizing and successfully shading required regions.
Accurately plotting boundary lines and curves relies on a good grasp of how to translate algebraic expressions into visual representations. Especially with complex functions like cubic ones, having a precise graph ensures clarity and correctness in finding shaded regions and intersections.

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Most popular questions from this chapter

Solve each problem. Tire Sales The number of automobile tire sales is dependent on several variables. In one study the relationship among annual tire sales \(S\) (in thousands of dollars), automobile registrations \(R\) (in millions), and personal disposable income \(I\) (in millions of dollars) was investigated. The results for three years are given in the table. To describe the relationship among these variables, we can use the equation $$ S=a+b R+c l $$ where the coefficients \(a, b,\) and \(c\) are constants that must be determined before the equation can be used. (Source: Jarrett, J., Business Forecasting Methods, Basil Blackwell, Ltd.) (a) Substitute the values for \(S, R,\) and \(I\) for each year from the table into the equation \(S=a+b R+c I,\) and obtain three linear equations involving \(a, b,\) and \(c\) (b) Use a graphing calculator to solve this linear system for \(a, b,\) and \(c .\) Use matrix inverse methods. (c) Write the equation for \(S\) using these values for the coefficients. (d) If \(R=117.6\) and \(I=310.73,\) predict \(S .\) (The actual value for \(S\) was \(11,314 .\) ) (e) If \(R=143.8 \text { and } I=829.06, \text { predict } S . \text { (The actual value for } S \text { was } 18,481 .)\) $$\begin{array}{|c|c|c|} \hline S & R & I \\ \hline 10,170 & 112.9 & 307.5 \\\ \hline 15,305 & 132.9 & 621.63 \\ \hline 21,289 & 155.2 & 1937.13 \\\ \hline \end{array}$$

Solve each problem. A glue company needs to make some glue that it can sell for \(120\) per barrel. It wants to use 150 barrels of glue worth \(100\) per barrel, along with some glue worth \(150\) per barrel and some glue worth \(190\) per barrel. It must use the same number of barrels of \(150\) and \(190\) glue. How much of the \(150\) and \(190\) glue will be needed? How many barrels of \(120\) glue will be produced?

Use Cramer's rule to solve each system of equations. If \(D=0,\) use another method to determine the solution set. $$\begin{aligned} -\frac{3}{4} x+\frac{2}{3} y &=16 \\ \frac{5}{2} x+\frac{1}{2} y &=-37 \end{aligned}$$

Use Cramer's rule to solve each system of equations. If \(D=0,\) use another method to determine the solution set. $$\begin{array}{r} 4 x+3 y=9 \\ 12 x+9 y=27 \end{array}$$

Given \(A=\left[\begin{array}{rr}4 & -2 \\ 3 & 1\end{array}\right], B=\left[\begin{array}{rr}5 & 1 \\ 0 & -2 \\ 3 & 7\end{array}\right],\) and \(C=\left[\begin{array}{rrr}-5 & 4 & 1 \\ 0 & 3 & 6\end{array}\right],\) find each product when possible. $$C B$$

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