Problem 35 - Algebra and Trigonometry with Analytic Geometry

Question

Is there a real number \(x\) such that \(x=2^{-x}\) ? Decide by displaying graphically the system $$ \left\\{\begin{array}{l} y=x \\ y=2^{-x} \end{array}\right. $$

Step-by-Step Solution

Answer

Yes, there is an intersection point where \(x=2^{-x}\), approximately \(x = 0.5\).

Step 1: Understand the Problem

We need to determine if there is a real number \(x\) such that \(x=2^{-x}\). This involves finding the intersection point(s) between the line \(y=x\) and the curve \(y=2^{-x}\).

Step 2: Sketch the Linear Function

First, we plot the line \(y=x\). This is a straight line that passes through the origin (0,0) with a slope of 1, which means for every unit increase in \(x\), \(y\) also increases by one unit.

Step 3: Sketch the Exponential Function

Now, plot the curve \(y=2^{-x}\). This is an exponential decay function. When \(x=0\), \(y=2^{0}=1\). As \(x\) increases, \(y\) approaches zero but never actually reaches it. As \(x\) becomes large and negative, \(y\) increases sharply.

Step 4: Find the Intersection Points

To find if \(x=2^{-x}\) has a real solution, we look for the intersection of the two graphs. By observing the plots of \(y=x\) and \(y=2^{-x}\), notice that they intersect at a single point.

Step 5: Verify the Intersection Point

From the graphical representation, the functions intersect somewhere between \(x=0\) and \(x=1\). To confirm this, pick a value like \(x=0.5\). Substitute in \(y=2^{-0.5}\), which gives \(0.5=2^{-0.5}\). This holds because \(2^{-0.5}=\frac{1}{\sqrt{2}}\approx0.707\). Thus, \(x=0.5\) is close to where the intersection occurs.

Key Concepts

Exponential Functions

An exponential function is a mathematical function of the form \(y = a^x\), where \(a\) is a constant and \(x\) is the exponent. In this context, the function \(y = 2^{-x}\) is called an exponential decay function. When the exponent is negative, it represents a function that decreases rapidly as \(x\) increases.

• When \(x = 0\), we have \(2^{-0} = 1\). This gives us the point (0, 1) on the graph, indicating that the function starts at 1 when \(x\) is zero.
• As \(x\) increases, \(y\) rapidly approaches zero, but never quite gets there. Thus, the x-axis serves as a horizontal asymptote.
• If \(x\) is negative, each increase makes \(y\) grow exponentially larger.

Understanding these features helps visualize where it might intersect with other functions, such as linear functions.

Linear Functions

A linear function is simply one that creates a straight line when graphed. The most basic form is \(y = mx + b\), with \(m\) as the slope and \(b\) as the y-intercept. In the problem, the linear function given is \(y = x\), which has a slope \(m = 1\) and a y-intercept \(b = 0\).

• This function creates a 45-degree line that passes through the origin (0,0).
• For every 1-unit increase in \(x\), \(y\) increases by the same amount, meaning the slope is 1.
• This kind of function is fundamental in mathematics because of its simplicity and predictability.

The line \(y = x\) serves as a reference point to explore intersections with more complex functions like exponential functions.

Intersection Points

The intersection point between two functions is where they share the same coordinates on a graph. This problem focuses on finding where the linear function \(y = x\) intersects the exponential function \(y = 2^{-x}\).

• Graphically, these appear as the points where the plotted lines cross each other.
• To determine the intersection algebraically, set the two equations equal: \(x = 2^{-x}\).
• Solving this equation involves using both graphing and algebraic techniques to find approximate or exact solutions.

Visualization is crucial here, as plotting the graphs provides a clear insight into where and if these functions intersect.

Real Solutions

A real solution is a value for \(x\) that satisfies both equations in the system with real numbers only. When graphing \(y = x\) and \(y = 2^{-x}\), we look for intersection points that can be confirmed visually or algebraically.

• In the graphical approach, we notice if and where the graphs meet. From the given task, the intersection occurs near \(x = 0.5\).
• This is verified by substituting into the equations to see if the values hold true, for example, \(2^{-0.5}\approx 0.707\).
• Real solutions imply those solutions have tangible, positive or negative real-number values, unlike imaginary numbers.

Confirming real solutions gives the assurance that the values are practical and applicable to real-world problems.