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Graph the solution set of each system of inequalities by hand. $$\begin{aligned} &y \leq \log x\\\ &y \geq|x-2| \end{aligned}$$

Short Answer

Expert verified
The solution is the area below \(y = \log x\) and above \(y = |x-2|\) on the graph.

Step by step solution

01

Graph the Boundary of the First Inequality

The inequality \(y \leq \log x\) has the boundary equation \(y = \log x\). Start by plotting the curve of the function \(y = \log x\). This curve has a vertical asymptote at \(x = 0\) and passes through points such as \((1, 0)\) and \((10, 1)\). The curve is increasing and lies in the first quadrant.
02

Shade the Solution Region for the First Inequality

Since the inequality is \(y \leq \log x\), shade the area below the curve \(y = \log x\) on your graph, including the points on the curve itself.
03

Graph the Boundary of the Second Inequality

For the second inequality \(y \geq |x-2|\), the boundary is \(y = |x-2|\). This absolute value function forms a 'V' shape with a vertex at \((2, 0)\). The lines \(y = x-2\) for \(x \geq 2\) and \(y = -(x-2)\) for \(x < 2\) form the 'V' shape. Plot this function on your graph.
04

Shade the Solution Region for the Second Inequality

Since the inequality is \(y \geq |x-2|\), shade the area above the 'V' formed by \(y = |x-2|\), including the lines themselves in the shading.
05

Identify the Solution Set of the System

The solution to the system of inequalities is the overlapping region from steps 2 and 4. This is the region that is shaded by both inequalities. Highlight this region on your graph.

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Key Concepts

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

Graphing Inequalities
Graphing inequalities involves representing solutions on a coordinate plane. When you graph an inequality like \(y \leq \log x\), you first graph the boundary equation, \(y = \log x\). This shows you the curve that serves as the limit for your inequality. Remember that logarithmic functions have a vertical asymptote at \(x = 0\), which means the graph will never actually touch or cross this line. This helps you to visualize the boundary effectively.
Once the boundary is plotted, the inequality \(\leq\) indicates that the solution set includes the area below the boundary line. Use shading techniques to denote this entire region on your graph. Similarly, the inequality \(y \geq |x-2|\) involves graphing the absolute value function \(y = |x-2|\), a 'V'-shaped graph. Again, shading above this 'V'-line helps identify the solution region, including the boundary itself.
  • Graphs are done on a coordinate plane.
  • Familiarize yourself with the shape of each function involved.
  • Shade the correct regions depending on the inequality symbols.
Understanding these concepts helps in visualizing how different regions on the graph represent solutions to the inequalities.
Logarithmic Functions
Logarithmic functions are a fundamental part of algebra and are often encountered in graphing exercises involving inequalities. The function \(y = \log x\) is a classic example of a logarithmic function, showing the relationship where \(y\) represents the power to which a base (commonly 10 in base-10 logarithms) must be raised to yield \(x\). This function is only defined for positive values of \(x\), and it features a vertical asymptote at \(x = 0\).
When graphing the logarithmic function \(y = \log x\), recognize some key characteristics:
  • The graph passes through the point \( (1, 0) \), since \(\log(1) = 0\).
  • It gradually increases, moving to the right as \(x\) increases.
  • The curve is never horizontal and doesn't touch the vertical asymptote.
This understanding is crucial when solving inequalities such as \(y \leq \log x\). Recognize that your shading should include the points on the graph where \(y\) is equal to \(\log x\) and below, showcasing a comprehensive solution.
Absolute Value Functions
Absolute value functions, such as \(y = |x-2|\), create a distinct 'V' shape on a graph. This type of graph is pivotal when studying systems of inequalities, because it demonstrates different behaviors depending on the value of \(x\).
The vertex of the 'V' is at \( (2, 0) \), which is the point where the graph changes direction. For \(x \geq 2\), the function follows the line \(y = x-2\), and for \(x < 2\), the graph mirrors this over the y-axis, following \(y = -(x-2)\).
  • The vertex represents the minimum point of the graph.
  • The graph is symmetric around the vertex.
In a situation where you graph \(y \geq |x-2|\), it signifies shading the region above and including the 'V' itself. This includes all points that satisfy the condition \(y = |x-2|\) and those above it, portraying the full solution set for the inequality. Understanding how to navigate through absolute value functions is key to solving and graphing more complex systems of inequalities efficiently.

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Most popular questions from this chapter

To analyze population dynamics of the northern spotted owl, mathematical ecologists divided the female owl population into three categories: juvenile (up to 1 year old), subadult (1 to 2 years old), and adult (over 2 years old). They concluded that the change in the makeup of the northern spotted owl population in successive years could be described by the following matrix equation. $$\left[\begin{array}{c} j_{n+1} \\ s_{n+1} \\ a_{n+1} \end{array}\right]=\left[\begin{array}{rrr} 0 & 0 & 0.33 \\ 0.18 & 0 & 0 \\ 0 & 0.71 & 0.94 \end{array}\right]\left[\begin{array}{c} j_{n} \\ s_{n} \\ a_{n} \end{array}\right]$$ The numbers in the column matrices give the numbers of females in the three age groups after \(n\) years and \(n+1\) years. Multiplying the matrices yields the following. $$\begin{aligned} &j_{n+1}=0.33 a_{n}\\\ &s_{n+1}=0.18 j_{n}\\\ &a_{n+1}=0.71 s_{n}+0.94 a_{n} \end{aligned}$$ (Source: Lamberson, R. H., R. McKelvey, B. R. Noon, and C. Voss, "A Dynamic Analysis of Northern Spotted Owl Viability in a Fragmented Forest Landscape," Conservation Biology, Vol. \(6, \text { No. } 4 .)\) (a) Suppose there are currently 3000 female northern spotted owls: 690 juveniles, 210 subadults, and 2100 adults. Use the preceding matrix equation to determine the total number of female owls for each of the next 5 years. (b) Using advanced techniques from linear algebra, we can show that, in the long run, $$ \left[\begin{array}{c} j_{n+1} \\ s_{n+1} \\ a_{n+1} \end{array}\right]=0.98359\left[\begin{array}{c} j_{n} \\ s_{n} \\ a_{n} \end{array}\right] $$ What can we conclude about the long-term fate of the northern spotted owl? (c) In this model, the main impediment to the survival of the northern spotted owl is the number 0.18 in the second row of the \(3 \times 3\) matrix. This number is low for two reasons: The first year of life is precarious for most animals living in the wild, and juvenile owls must eventually leave the nest and establish their own territory. If much of the forest near their original home has been cleared, then they are vulnerable to predators while searching for a new home. Suppose that, due to better forest management, the number 0.18 can be increased to \(0.3 .\) Rework part (a) under this new assumption.

As the price of a product increases, businesses usually increase the quantity manufactured. However, as the price increases, consumer demand-or the quantity of the product purchased by consumers-usually decreases. The price we see in the market place occurs when the quantity supplied and the quantity demanded are equal. This price is called the equilibrium price and this demand is called the equilibrium demand. The supply of a certain product is related to its price by the equation \(p=\frac{1}{3} q,\) where \(p\) is in dollars and \(q\) is the quantity supplied in hundreds of units. (a) If this product sells for 9 dollars, what quantity will be supplied by the manufacturer? (b) Suppose that consumer demand for the same product decreases as price increases according to the equation \(p=20-\frac{1}{5} q .\) If this product sells for 9 dollars, what quantity will consumers purchase? How does this compare with the quantity being supplied by the manufacturer at this price? (c) On the basis of parts (a) and (b), what should happen to the price? Explain. (d) Determine the equilibrium price at which the quantity supplied and quantity demanded are equal. What is the demand at this price?

Solve each system graphically. Check your solutions. Do not use a calculator. $$\begin{aligned}&x-y=3\\\&x^{2}+y^{2}=9\end{aligned}$$

Because variables appear in denominators, the system $$\begin{aligned}&\frac{5}{x}+\frac{15}{y}=16\\\&\frac{5}{x}+\frac{4}{y}=5\end{aligned}$$ is not a linear system. However, we can solve it in a manner similar to the method for solving a linear system by using a substitution-of-variable technique. Let \(t=\frac{1}{x}\) and let \(u=\frac{1}{y} \). Solve the given system for \(x\) and \(y\) by using the equations relating \(t\) to \(x\) and \(u\) to \(y\).

Let \(A=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right], B=\left[\begin{array}{ll}b_{11} & b_{12} \\ b_{21} & b_{22}\end{array}\right],\) and \(C=\left[\begin{array}{ll}c_{11} & c_{12} \\\ c_{21} & c_{22}\end{array}\right]\) where all the elements are real numbers. Use these matrices to show that each statement is true for \(2 \times 2\) matrices. \((A B) C=A(B C)\) (associative property)

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