Skip to main content
Ch. 5 - Systems of Equations and Inequalities
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
All textbooksBlitzer 8th EditionCh. 5 - Systems of Equations and InequalitiesProblem 25
Chapter 6, Problem 25

Graph each inequality. y≥log2(x+1)

Verified step by step guidance
1
Identify the inequality to graph: \(y \geq \log_{2}(x+1)\). This means we are looking at all points \((x,y)\) where \(y\) is greater than or equal to the logarithm base 2 of \((x+1)\).
Determine the domain of the function \(y = \log_{2}(x+1)\). Since the logarithm is defined only for positive arguments, set \(x+1 > 0\), which gives \(x > -1\). So, the graph will only exist for \(x > -1\).
Graph the boundary curve \(y = \log_{2}(x+1)\). This is the logarithmic function shifted left by 1 unit. Plot key points such as when \(x=0\), \(y=\log_{2}(1)=0\), and when \(x=3\), \(y=\log_{2}(4)=2\), to help sketch the curve.
Since the inequality is \(y \geq \log_{2}(x+1)\), shade the region above or on the curve. This includes the curve itself because of the 'equal to' part.
Draw a solid line for the curve \(y = \log_{2}(x+1)\) to indicate that points on the curve satisfy the inequality, and shade the area above this curve for all \(x > -1\).

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
6m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Logarithmic Functions

A logarithmic function is the inverse of an exponential function. For example, y = log₂(x + 1) means y is the power to which 2 must be raised to get (x + 1). Understanding the domain and range of logarithmic functions is essential for graphing them correctly.
Recommended video:
5:26
Graphs of Logarithmic Functions

Graphing Inequalities

Graphing inequalities involves shading the region of the coordinate plane that satisfies the inequality. For y ≥ log₂(x + 1), you first graph y = log₂(x + 1) as a boundary curve, then shade the area above or on the curve where y-values are greater than or equal to the logarithmic function.
Recommended video:
Guided course
7:02
Linear Inequalities

Domain Restrictions

The domain of y = log₂(x + 1) is x > -1 because the argument of a logarithm must be positive. Recognizing domain restrictions helps in accurately plotting the graph and understanding where the inequality is defined.
Recommended video:
3:51
Domain Restrictions of Composed Functions
Related Practice
Textbook Question

In Exercises 19–28, solve each system by the addition method. {y2−x=4x2+y2=4\(\begin{cases}\)y^2 - x = 4 \(\x\)^2 + y^2 = 4\(\end{cases}\)

4
views
Textbook Question

In Exercises 19–30, solve each system by the addition method. 4x + 3y = 15 2x - 5y = 1

Textbook Question

In Exercises 25–35, solve each system by the method of your choice. This is a piecewise function, refer to textbook problem.

5
views
Textbook Question

Write the partial fraction decomposition of each rational expression. x2+2x+7/x(x − 1)2

3
views
Textbook Question

In Exercises 19–28, solve each system by the addition method. {x2+y2=25(x−8)2+y2=41\(\begin{cases}\)x^2 + y^2 = 25 \\(x - 8)^2 + y^2 = 41\(\end{cases}\)

Textbook Question

Solve each system in Exercises 25–26. {x+32−y−12+z+24=32x−52+y+13−z4=−256x−34−y+12+z−32=−52\(\begin{cases}\[\frac{x + 3}{2}\) - \(\frac{y - 1}{2}\) + \(\frac{z + 2}{4}\) = \(\frac{3}{2}\) \(\frac{x - 5}{2}\) + \(\frac{y + 1}{3}\) - \(\frac{z}{4}\) = - \(\frac{25}{6}\) \(\frac{x - 3}{4}\) - \(\frac{y + 1}{2}\) + \(\frac{z - 3}{2}\) = - \(\frac{5}{2}\]\end{cases}\)

3
views