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Chapter 10: Problem 12

Graph the given inequalities. $$y \leq \sqrt{x}$$

Short Answer

Expert verified
Shade below the curve \(y = \sqrt{x}\) including points on it.

Step by step solution

01

Understand the Inequality

The inequality given is \( y \leq \sqrt{x} \). This represents all the points on or below the curve \( y = \sqrt{x} \), where \( y \) is the dependent variable and \( x \) is the independent variable.
02

Sketch the Basic Function

First, sketch the function \( y = \sqrt{x} \). This is the upper half of a parabola opening to the right. It passes through points like (0,0), (1,1), (4,2), and (9,3). Plot these points on the coordinate plane.
03

Draw the Boundary Line

Since \( y \leq \sqrt{x} \) includes the boundary, draw the curve \( y = \sqrt{x} \) as a solid line. This indicates that points on the line are included in the solution set.
04

Shade the Solution Region

Shade the region below the curve \( y = \sqrt{x} \) to show the solution set for the inequality \( y \leq \sqrt{x} \). Include the region that extends infinitely to the left since any point with a negative \( x \) would not make sense in \( y = \sqrt{x} \), as this function starts at \( x = 0 \).
05

Verify Your Graph

Check several points to ensure they satisfy the inequality. For example, at (1,0), since \( 0 \leq \sqrt{1} \), this point is in the solution. Similarly, verify other points if needed.

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Key Concepts

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

Coordinate Plane
A coordinate plane is a foundational concept for graphing any function or inequality. It is a two-dimensional plane formed by the intersection of a vertical y-axis and a horizontal x-axis.
The plane is divided into four quadrants, but for the inequality \( y \leq \sqrt{x} \), we are mostly interested in the first quadrant, where both x and y are positive.
Important features of the coordinate plane include:
  • Origin: The point \((0,0)\) where the x-axis and y-axis intersect.
  • Quadrants: The plane is divided into four sections, but in this exercise, the relevant area is typically the first quadrant.
  • Axes: The horizontal axis (x-axis) and the vertical axis (y-axis) help locate points.
When graphing, always ensure the interval you're working with is labeled for clarity and correctness. Plotting specific points helps visualize the solution to an inequality.
Function Sketching
Function sketching is an essential skill when dealing with inequalities. In this problem, we need to sketch the curve \( y = \sqrt{x} \), which represents the boundary for our inequality.
Steps for sketching:
  • Determine the shape: \( y = \sqrt{x} \) is part of a parabola that opens to the right.
  • Identify key points: Plot points such as \( (0,0), (1,1), (4,2), (9,3) \). These help guide the sketch of the curve.
  • Connect the dots: Draw a smooth curve through these points to get the correct shape of \( y = \sqrt{x} \).
Sketching functions accurately sets the stage for exploring inequalities, providing a clear view of which regions of the plane satisfy the inequality's conditions.
Inequality Solution
Solving the inequality \( y \leq \sqrt{x} \) involves identifying all the sets of points that make the inequality true. An important concept here is inequality representation graphically.
For \( y \leq \sqrt{x} \), we need to include all the points on the curve and below it.
Here's how we solve the inequality:
  • On the curve: Points on \( y = \sqrt{x} \) satisfy the \( \leq \) part.
  • Below the curve: Any point \((x,y)\) where \( y < \sqrt{x} \) is a part of the solution.
  • Boundary inclusion: Use a solid line to indicate that points on the curve are included.
This graphically shows where the inequality holds true, marking a region rather than just a straight line, highlighting the difference between equations and inequalities.
Shading Regions
Shading the appropriate region of the graph confirms the solution set for an inequality visually. For \( y \leq \sqrt{x} \), shading the region below the curve provides an instant visual cue to the solution.
Here鈥檚 how to shade the correct area:
  • Determine the boundary: Since points on \( y = \sqrt{x} \) are included, the line itself should not be skipped.
  • Identify the correct region: Test points below the curve ensure they satisfy the inequality.
  • Extend shading: Ensure the shaded area covers all relevant points, including along the x-axis from the origin.
Shading effectively distinguishes the region that meets the inequality criteria from areas that do not, offering quick verification of its truth across different parts of the coordinate plane.

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Most popular questions from this chapter

A function \(f\) is defined as follows. The domain of \(f\) is the set of all \(2 \times 2\) matrices (with real entries). If \(A=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)\) then \(f(A)=a d-b c\) (a) Let \(A=\left(\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right),\) and \(B=\left(\begin{array}{cc}3 & -1 \\ 5 & 8\end{array}\right) .\) Compute \(f(A), f(B),\) and \(f(A B) .\) Is it true, in this case, that \(f(A) \cdot f(B)=f(A B) ?\) (b) Let \(A=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)\) and \(B=\left(\begin{array}{ll}e & f \\ g & h\end{array}\right) .\) Show that \(f(A) \cdot f(B)=f(A B)\).

In this exercise, let's agree to write the coordinates \((x, y)\) of a point in the plane as the \(2 \times 1\) matrix \(\left(\begin{array}{l}x \\\ y\end{array}\right) .\) (a) Let \(A=\left(\begin{array}{rr}1 & 0 \\ 0 & -1\end{array}\right)\) and \(Z=\left(\begin{array}{l}x \\ y\end{array}\right) .\) Compute the ma- trix \(A Z .\) After computing \(A Z,\) observe that it represents the point obtained by reflecting \(\left(\begin{array}{l}x \\ y\end{array}\right)\) about the \(x\) -axis. (b) Let \(B=\left(\begin{array}{rr}-1 & 0 \\ 0 & 1\end{array}\right)\) and \(Z=\left(\begin{array}{l}x \\ y\end{array}\right) .\) Compute the ma- trix \(B Z\). After computing \(B Z\), observe that it represents the point obtained by reflecting \(\left(\begin{array}{l}x \\ y\end{array}\right)\) about the \(y\) -axis. (c) Let \(A, B,\) and \(Z\) represent the matrices defined in parts \((a)\) and \((b) .\) Compute the matrix \((A B) Z\) and then interpret it in terms of reflection about the axes.

Use Cramers rule to solve those systems for which \(D \neq 0 .\) In cases where \(D=0,\) use Gaussian elimination or matrix methods. $$\left\\{\begin{array}{l} 5 x-3 y-z=16 \\ 2 x+y-3 z=5 \\ 3 x-2 y+2 z=5 \end{array}\right.$$

Let \(a, b,\) and \(c\) be constants (with \(a \neq 0\) ), and consider the system $$\left\\{\begin{array}{l}y=a x^{2}+b x+c \\\y=k\end{array}\right.$$ For which value of \(k\) (in terms of \(a, b,\) and \(c\) ) will the system have exactly one solution? What is that solution? What is the relationship between the solution you've found and the graph of \(y=a x^{2}+b x+c ?\)

The matrices \(A, B, C, D, E, F,\) and \(G\) are defined as follows: $$ A=\left(\begin{array}{rr} 2 & 3 \\ -1 & 4 \end{array}\right) \quad B=\left(\begin{array}{rr} 1 & -1 \\ 3 & 0 \end{array}\right) \quad C=\left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right) $$ $$\begin{aligned} &D=\left(\begin{array}{rrr} -1 & 2 & 3 \\ 4 & 0 & 5 \end{array}\right) \quad E=\left(\begin{array}{rr} 2 & 1 \\ 8 & -1 \\ 6 & 5 \end{array}\right)\\\ &F=\left(\begin{array}{rr} 5 & -1 \\ -4 & 0 \\ 2 & 3 \end{array}\right) \quad G=\left(\begin{array}{ll} 0 & 0 \\ 0 & 0 \\ 0 & 0 \end{array}\right) \end{aligned}$$ In each exercise, carry out the indicated matrix operations if they are defined. If an operation is not defined, say so. $$5 E-3 F$$
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