Problem 8 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Graph \(f(x)=e^{x}-x\). Only use a calculator to check your work after working on your own. (a) Find \(f^{\prime}(x)\). Draw a number line and indicate where \(f^{\prime}\) is positive, zero, and negative. (b) Label the \(x\) - and \(y\) -coordinates of any local extrema (local maxima or minima). (c) Using your picture, determine how many solutions there are to the following equations. i. \(f(x)=5\) ii. \(f(x)=0.5\) Notice that these equations are "intractable" - try to solve \(e^{x}-x=5\) algebraically to see what this means. If we want to estimate the solutions, we can do so using a graphing calculator. At this point, we should know how many solutions to expect.

Step-by-Step Solution

Answer

Derivative of \(f(x)=e^x-x\) is \(f'(x)=e^x-1\). The derivative is zero at \(x=0\), negative for \(x<0\), and positive for \(x>0\). The local minimum is at (0,1). There is one solution for \(f(x)=5\) and two solutions for \(f(x)=0.5\).

Step 1: Find The Derivative

First, find the derivative of \(f(x)=e^x-x\), which is \(f'(x)=e^x-1\). This will be used later to determine where the function is increasing and decreasing.

Step 2: Identify Positive, Zero and Negative Values of f'(x)

Next, determine where \(f'(x) = 0\) by solving \(e^x - 1 = 0\). Note that the function \(e^x\) is always positive, the only value possible to achieve 0 is when \(x = 0\). When \(x < 0\), \(e^x < 1\), thus \(f'(x) < 0\). For \(x > 0\), \(e^x > 1\), thus \(f'(x) > 0\). This implies that there is a turning point at \(x = 0\).

Step 3: Identify Local Extrema

As there is a turning point at \(x=0\), this is where the function changes from decreasing to increasing. Thus, \(x = 0\) is the x-coordinate for a local minimum. To find the y-coordinate for the local minimum, substitute \(x = 0\) into \(f(x) = e^x - x\), yielding \(f(0) = e^0 - 0 = 1\). Therefore, the local minimum is at (0, 1).

Step 4: Estimate the Solution

From the graph, estimate the \(x\) values where \(f(x) = 5\) and \(f(x) = 0.5\). Note that these values cannot be solved algebraically. Given the function's behavior at \(x = 0\) and as \(x\) approaches \(+\inf\) and \(-\inf\), for \(f(x) = 5\), there should be one solution; for \(f(x) = 0.5\), there are two solutions. This can be confirmed by checking the graph.

Key Concepts

Derivative Calculation

When we talk about derivatives in calculus, we're discussing how a function changes at any given point. A derivative represents the slope of the tangent line at a particular point on a graph of a function. In essence, it's the instantaneous rate of change. The calculation of a derivative begins with a function, in this case, we have the function f(x) = e^x - x. To find the derivative, denoted as f'(x), we apply rules of differentiation.

For the exponential part, e^x, the derivative remains the same due to the unique nature of the exponential function. For the linear part, -x, the derivative is simply -1 since the slope of x is 1 and we're subtracting it. Combining these, we get f'(x) = e^x - 1. Understanding this step is fundamental, as the derivative will guide us in finding where the function increases, decreases, and has any possible local extrema.

Local Extrema Identification

Identifying local extrema—that is, the peaks (maxima) and troughs (minima) on a graph—is key to understanding the function's overall behavior. We typically find these turning points by setting the derivative equal to zero and solving for x. In our case, solving f'(x) = e^x - 1 = 0 leads us to the value x = 0.

Why is this significant? A derivative that changes sign around a point suggests a local extremum. Here, the derivative changes from negative to positive as x increases through zero, indicating the existence of a local minimum at this point. To find the precise coordinates of this local minimum, we substitute x = 0 back into our original function, hence f(0) = e^0 - 0 = 1. Thus, our local minimum is at the point (0, 1). Knowing local extrema is pivotal for sketching the graph of the function and for understanding its geometry and real-world applications.

Graphical Solution Estimation

Sometimes, equations are so complex that they cannot be solved with simple algebra and require estimation techniques. This is where the concept of graphical solution estimation comes in. By graphing the function f(x) = e^x - x, we can visually estimate the solutions to equations such as f(x) = 5 and f(x) = 0.5.

Using a graphing calculator can be immensely helpful for this purpose, but we should first have a mental image of the graph based on the function’s behavior. For instance, since we know that the function has a local minimum at (0, 1) and we understand the behavior of both e^x and -x independently, we can anticipate how the function behaves as x approaches infinity or negative infinity. When estimating solutions, it's crucial to consider how many times the graph will intersect the horizontal lines y = 5 and y = 0.5. Based on the function's increase and decrease intervals, we can infer the number of solutions without explicitly solving the equations.