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Use a graphing utility to solve the system of equations. Find the solution(s) accurate to two decimal places. $$\left\\{\begin{aligned} y &=e^{x} \\ x-y+1 &=0 \end{aligned}\right.$$

Short Answer

Expert verified
The estimated solution point from the graph will be the accurate to two decimal places value where the graphs of both functions intersect.

Step by step solution

01

Plot the Functions

Choose a standard window that shows the intersection point well, and plot \(y = e^x\) and \(x - y + 1 = 0\) or equivalently \(y = x + 1\) using a graphing calculator or online tool.
02

Identify Intersection

Identify the point where the two graphs intersect. This point constitutes the solution to the system of equations.
03

Finding Approximate Values

Hover over or click on the intersection point to view the coordinates. Round these according to the required decimal places. For our case, it's two decimal places.

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Key Concepts

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

Graphing Utility
Using a graphing utility is a straightforward way to visualize mathematical functions and solve equations. Graphing utilities can include software applications like graphing calculators, online tools, or computer software programs. They allow you to input equations and see their representations on a coordinate plane. This visual aid helps in understanding relationships between variables in equations easily.

To use a graphing utility effectively, follow these steps:
  • Input the equations: Enter the functions you want to graph. In the exercise, these are: \(y = e^x\) and \(y = x + 1\).
  • Set a suitable window: Adjust the viewing window on the graphing utility so that you can view the behavior of the functions clearly and locate points of interest like intersections.
  • Plot the graphs: Once the functions and window are set, plot the graphs and observe their shapes and the regions they occupy on the plane.
The power of graphing utilities lies in their capacity to solve systems of equations visually, by identifying intersection points directly on the graph.
Intersection of Graphs
The intersection of graphs is a critical concept in understanding systems of equations. This is where two or more graphs meet on the coordinate plane, representing the point(s) where the systems of equations have equal values.

In our problem, the function \(y = e^x\) and the line \(y = x + 1\) need to be plotted. The point where these two graphs intersect is the solution to the system of equations. Mathematically, an intersection means the x-value and y-value are the same for both functions at that particular point.

To find this intersection:
  • Observe the plotted graphs on the utility.
  • Locate the exact point where they cross each other. This point is the solution of the system, giving you the x and y coordinates that satisfy both equations simultaneously.
  • Use the graphing tool's feature to accurately identify the intersection coordinates to the required precision, in this case, two decimal places.
Finding intersections graphically provides a clear visual solution to complex algebraic problems, making it easier to comprehend and interpret.
Exponential Functions
Exponential functions are equations in which a variable is raised to a constant power. The form is often written as \(y = a^x\), where \(a\) is a constant. The exponential function featured in the problem is \(y = e^x\), a specific case with the base \(e\), the Euler's number, approximately equal to 2.718.

Key characteristics of exponential functions include:
  • Rapid Growth: As \(x\) increases, \(e^x\) grows very quickly, resulting in a steep curve upward on a graph.
  • Horizontal Asymptote: As \(x\) approaches negative infinity, \(e^x\) approaches zero but never actually reaches it.
  • Continuous and Smooth Curve: The graph of \(e^x\) is smooth with no breaks, featuring a continuous upward trend.
When solving a system involving an exponential function like \(y = e^x\), graphing is particularly helpful. Graphical methods provide insights into how the exponential function behaves in comparison with other types, such as linear functions, as seen in the example \(y = x + 1\). Recognizing the distinct features of exponentials assists in anticipating their intersection points with linear functions, as demonstrated through their respective graphical plots.

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Most popular questions from this chapter

Acid Mixture A chemist needs 10 liters of a \(25 \%\) acid solution. The solution is to be mixed from three solutions whose concentrations are \(10 \%, 20 \%,\) and \(50\%\). How many liters of each solution will satisfy each condition? (a) Use 2 liters of the \(50 \%\) solution. (b) Use as little as possible of the \(50 \%\) solution. (c) Use as much as possible of the \(50 \%\) solution.

Find the equation of the parabola $$y=a x^{2}+b x+c$$ that passes through the points. To verify your result, use a graphing utility to plot the points and graph the parabola. $$(0,0),(2,-2),(4,0)$$

Sketch the region determined by the constraints. Then find the minimum and maximum values of the objective function (if possible) and where they occur, subject to the indicated constraints. Objective function: \(z=3 x+2 y\) Constraints: $$\begin{array}{r}x \geq 0 \\\y \geq 0 \\\5 x+2 y \leq 20 \\\5 x+y \geq 10\end{array}$$

Sketch the region determined by the constraints. Then find the minimum and maximum values of the objective function (if possible) and where they occur, subject to the indicated constraints. Objective function: \(z=5 x+\frac{1}{2} y\) Constraints: $$\begin{aligned}x & \geq 0 \\ y & \geq 0 \\\\\frac{1}{2} x+y & \leq 8 \\\x+\frac{1}{2} y & \geq 4\end{aligned}$$

Determine whether the statement is true or false. Justify your answer. When solving a linear programming problem, if the objective function has a maximum value at more than one vertex, then there are an infinite number of points that will produce the maximum value.

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