Problem 45 - Precalculus with Calculus Previews

Question

At the beginning of this section we saw that the derivative of \(f(x)=e^{x}\) is \(f^{\prime}(x)=e^{x} .\) Use this information to find all tangent lines to the graph of \(f(x)=e^{x}\) that pass through the origin.

Step-by-Step Solution

Answer

The tangent line is \(y = ex\).

Step 1: Understand the Given Information

We are given the function \(f(x) = e^x\) and its derivative \(f'(x) = e^x\). The problem asks for the tangent lines to this curve that pass through the origin (0,0).

Step 2: Use the Point-Slope Form of a Line

The equation of a line in point-slope form is \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is a point on the line. Here, the point on the tangent line is \(x_1, e^{x_1}\) and the slope at this point is \(e^{x_1}\).

Step 3: Set the Condition for Tangent Line through the Origin

For the tangent line to pass through the origin, substitute \((0,0)\) into the point-slope equation: \(0 - e^{x_1} = e^{x_1}(0 - x_1)\). This simplifies to: \(e^{x_1} = e^{x_1}x_1\).

Step 4: Simplify the Equation

Divide both sides of \(e^{x_1} = e^{x_1}x_1\) by \(e^{x_1}\) to get: \(1 = x_1\). Thus, the point of tangency is \((1, e^1) = (1, e)\).

Step 5: Find the Equation of the Tangent Line

Use \(x_1 = 1\) and the point \((1, e)\) to write the tangent line using the point-slope form: \(y - e = e(x - 1)\), which simplifies to \(y = ex - e + e = ex\). So, the tangent line's equation is \(y = ex\).

Key Concepts

Derivative

The concept of a derivative is fundamental in calculus and describes how a function changes as its input changes. When we talk about the derivative of a function, we are essentially discussing the function's rate of change or its "slope" at any given point. For example, if you have a function \(f(x) = e^x\), the derivative, denoted as \(f'(x)\), is also \(e^x\). This tells us that the slope of the tangent line to the graph of \(f(x) = e^x\) at any point is \(e^x\) itself.

• The derivative is often called the "instantaneous rate of change".
• For linear functions, the derivative is constant and represents the slope of the line.
• For non-linear functions, the derivative can vary depending on the value of \(x\).

Understanding derivatives helps solve problems involving rates of change, such as velocity and acceleration in physics, or growth rates in biology. In the exercise, finding the derivative \(f'(x) = e^x\) enables determining the slope of tangents to \(f(x) = e^x\) that pass through a particular point: the origin in this case.

Exponential Function

An exponential function is a mathematical function of the form \(f(x) = a^{x}\), where \(a\) is a constant, and \(a > 0 \). One of the most well-known exponential functions is \(f(x) = e^x\), where \(e\) is Euler's number, approximately equal to 2.71828. Exponential functions describe a diversity of processes, such as population growth and radioactive decay.

• The function \(e^x\) grows rapidly as \(x\) increases because its rate of increase is proportional to its current value.
• The graph of \(e^x\) is always above the x-axis because exponential functions are never zero or negative for real numbers.
• This function is unique in calculus, as its derivative is the same as itself.

In the given exercise, the function \(f(x) = e^x\) is pivotal because of its simplicity and self-derivative nature. This makes it particularly interesting in problems involving tangents and graph behaviors.

Point-Slope Form

The point-slope form is a useful tool for finding the equation of a line when you know a point on the line and its slope. The general format is \(y - y_1 = m(x - x_1)\), where \(m\) denotes the slope, and \((x_1, y_1)\) is a point on the line. By rearranging this form, you can easily convert it into the slope-intercept format or other forms as needed.

• The point-slope form is derived from the basic definition of a slope \(m = \frac{{y_2 - y_1}}{{x_2 - x_1}}\).
• This form is especially helpful for fitting tangents to curves at given points.
• It simplifies to \(y = mx + b\) when \((x_1, y_1)\) is expanded, making it versatile in use.

In the step-by-step solution, substituting the origin point \((0,0)\) into the point-slope form ensures that any tangent line that touches \(f(x) = e^x\) at \(x_1\) and passes through the origin can be determined effectively.

Graph Analysis

Graph analysis involves examining the graph of a function to determine various characteristics such as intercepts, slopes, asymptotes, and behavior as \(x\) approaches infinity or negative infinity. When working with the graph of \(f(x) = e^x\), we can identify several important traits:

• The y-intercept occurs at (0,1) because \(e^0 = 1\).
• The x-axis acts as a horizontal asymptote, implying the graph approaches but never reaches zero as \(x\) decreases.
• Tangent lines have varying slopes across the graph since \(e^x\) changes exponentially.

In the exercise, discovering the tangent line that passes through the origin involves analyzing the graph to find the correct point of tangency by setting up conditions. We use mathematical techniques and derivative insights to determine that the tangent line at \(x = 1\) not only touches the curve but also extends through the origin, revealing geometric properties of \(e^x\).