Graph $f(x)=2^x+1$ and $g(x)=-x+4$ on the same coordinate plane.
What is the solution to the equation $f(x)=g(x)$ ?
We are tasked with finding the solution to the equation \(f(x) = g(x)\), where \(f(x) = 2^x + 1\) and \(g(x) = -x + 4\). Finding the solution to this equation is equivalent to finding the \(x\)-coordinate where the graphs of these two functions intersect on the Cartesian plane. The function \(f(x) = 2^x + 1\) is an exponential function that is always positive and strictly increasing as \(x\) increases. Conversely, the function \(g(x) = -x + 4\) is a linear function with a negative slope, meaning it is strictly decreasing as \(x\) increases. Because one function is strictly increasing and the other is strictly decreasing, they can intersect at most once.
We set the two expressions equal to each other to find the intersection point:
\[2^x + 1 = -x + 4\]
To simplify, we can move all terms to one side:
\[2^x + x - 3 = 0\]
We are looking for a value of \(x\) that satisfies this equation. Let's test simple integer values. If we substitute \(x = 1\):
\[2^1 + 1 = 3\]
\[-1 + 4 = 3\]
Since \(f(1) = 3\) and \(g(1) = 3\), we see that \(x = 1\) is a valid solution.
As established in the analysis, since \(f(x) = 2^x + 1\) is strictly increasing and \(g(x) = -x + 4\) is strictly decreasing, there cannot be any other intersection points. Any increase in \(x\) would cause \(f(x)\) to increase further away from its value at \(x=1\), while \(g(x)\) would decrease, moving further away as well. Thus, \(x = 1\) is the unique solution to the equation. The graphs of \(f(x)\) and \(g(x)\) intersect exactly at the point \((1, 3)\). Therefore, the solution to the equation \(f(x) = g(x)\) is \(x = 1\).
In physics and economics, finding the intersection of two functions often represents an equilibrium point. For instance, if \(f(x)\) represents the supply function of a good and \(g(x)\) represents the demand function, the intersection point \(f(x) = g(x)\) identifies the market equilibrium where the quantity supplied equals the quantity demanded at a specific price. Solving these systems algebraically, as done here, allows us to precisely determine these critical balancing points without needing to manually sketch every possible graph.