Problem 57
Question
graph f and g in the same rectangular coordinate system. Then find the point of intersection of the two graphs. $$ f(x)=2^{x}, g(x)=2^{-x} $$
Step-by-Step Solution
Verified Answer
The point of intersection between the two graphs is at (0, 1).
1Step 1: Graph the functions
To graph the equations, create a table of values for each function, \(f(x)\) and \(g(x)\). Select a few values for x and use this to compute the corresponding y-values. Then plot these points on the rectangular coordinate system and sketch the graph.
2Step 2: Identify potential intersection points
Examine the graphs and identify where the lines cross. These are the potential points of intersection. From the graph, we can see that the two graphs cross at x=0.
3Step 3: Solve for the point of intersection algebraically
To confirm the intersection point, you can solve the equation when \(f(x) = g(x)\). So, we have the equation \(2^{x} = 2^{-x}\). Since these are exponents with the same base, the rule is that if \(a^b = a^c\), then \(b = c\). So, \(2^{x} = 2^{-x}\) is true when \(x=-x\). Solving this gives \(x = 0\). Substitute \(x = 0\) into any of the original equations to find the corresponding y-value. Thus, f(0) = 2^0 = 1.
Key Concepts
Exponential FunctionsRectangular Coordinate SystemAlgebraic Solution
Exponential Functions
Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. They play a central role in representing situations where growth or decay accelerates rapidly—in other words, exponential changes.
In our exercise, the functions are given by \( f(x) = 2^x \) and \( g(x) = 2^{-x} \). Here's a quick breakdown:
In our exercise, the functions are given by \( f(x) = 2^x \) and \( g(x) = 2^{-x} \). Here's a quick breakdown:
- In \( f(x) = 2^x \), as \( x \) increases, the value of \( f(x) \) doubles; hence, it represents exponential growth.
- In \( g(x) = 2^{-x} \), as \( x \) increases, the value of \( g(x) \) halves; thus, it models exponential decay.
Rectangular Coordinate System
The rectangular coordinate system, also known as the Cartesian coordinate system, allows us to graphically represent equations. By plotting graphs of functions on the same set of axes, students can visually interpret how these functions behave and intersect.
When graphing an equation like \( f(x) = 2^x \) in this system, every point on the graph corresponds to a pair \((x, y)\), where \( x \) is any chosen input, and \( y \) is the resulting output. For example, if \( x \) is 0, then \( f(x) = 2^0 = 1 \), giving the point \((0, 1)\).
For both \( f(x) \) and \( g(x) \), plotting several points and connecting them allows us to visualize their behaviors. The intersection of their graphs on the same coordinate system shows where both functions share the same \( x, y \) values, indicating a solution to both equations simultaneously. This kind of graphing is critical for understanding relationships and solving problems in mathematics.
When graphing an equation like \( f(x) = 2^x \) in this system, every point on the graph corresponds to a pair \((x, y)\), where \( x \) is any chosen input, and \( y \) is the resulting output. For example, if \( x \) is 0, then \( f(x) = 2^0 = 1 \), giving the point \((0, 1)\).
For both \( f(x) \) and \( g(x) \), plotting several points and connecting them allows us to visualize their behaviors. The intersection of their graphs on the same coordinate system shows where both functions share the same \( x, y \) values, indicating a solution to both equations simultaneously. This kind of graphing is critical for understanding relationships and solving problems in mathematics.
Algebraic Solution
To find where two functions intersect, an algebraic solution involves setting their equations equal to each other and solving for \( x \). This method offers precision that graphing sometimes cannot provide.
In our exercise, to find the intersection point of \( f(x) = 2^x \) and \( g(x) = 2^{-x} \), we equate them: \[ 2^x = 2^{-x} \]Since they share the same base, we can equate their exponents:\[ x = -x \]Adding \( x \) to both sides gives:\[ 2x = 0 \]Thus, \( x = 0 \). Plugging this back into either function to find the \( y \)-value, we use \( f(0) = 2^0 = 1 \), confirming the intersection point at \((0, 1)\).
This algebraic method underlines the importance of understanding exponent rules and manipulation. It provides a way to corroborate graphical solutions and understand the deeper relationships between mathematical expressions.
In our exercise, to find the intersection point of \( f(x) = 2^x \) and \( g(x) = 2^{-x} \), we equate them: \[ 2^x = 2^{-x} \]Since they share the same base, we can equate their exponents:\[ x = -x \]Adding \( x \) to both sides gives:\[ 2x = 0 \]Thus, \( x = 0 \). Plugging this back into either function to find the \( y \)-value, we use \( f(0) = 2^0 = 1 \), confirming the intersection point at \((0, 1)\).
This algebraic method underlines the importance of understanding exponent rules and manipulation. It provides a way to corroborate graphical solutions and understand the deeper relationships between mathematical expressions.
Other exercises in this chapter
Problem 57
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