Problem 56 - Calculus of a Single Variable

Question

Finding an Equation of a Tangent Line In Exercises \(55-62,\) find an equation of the tangent line to the graph of the function at the given point. $$ f(x)=e^{-2 x}, \quad(0,1) $$

Step-by-Step Solution

Answer

The equation of the tangent line to the function \( f(x)=e^{-2x} \) at the point \( (0,1) \) is \( y=-2x+1 \).

Step 1: Find the Derivative

First, we need to find the derivative of \( f(x)=e^{-2x} \). Using the chain rule, we have \( f'(x) = -2e^{-2x} \).

Step 2: Find the slope of the tangent line

Next, we need to calculate the slope of the tangent line at the at \( x=0 \). As the slope of the tangent line at a point is given by the value of the derivative at that point, substituting \( x=0 \) in the derivative gives the slope \( f'(0) = -2e^{-2(0)} = -2 \).

Step 3: Find the equation of the tangent line

We now have the slope, \( m = -2 \), and a point, \( (0,1) \), on the line. We can substitute these values in the equation \( y = mx + b \) and solve for \( b \) to find the equation of the tangent line. The y-coordinate of the given point is equal to the y-intercept, hence \( b=1 \). This gives the equation of the tangent line as \( y=-2x+1 \).

Key Concepts

Derivative of Exponential Functions

Understanding the derivative of exponential functions is vital when working with the growth and decay models they often represent. Take for instance the function in our exercise, f(x) = e^{-2x}. The exponential function, e^x, is particularly special because its derivative is also e^x. However, when the exponential function has a coefficient in the exponent, such as -2x in our example, the derivative changes.

The process of finding the derivative of such functions involves a rule from calculus known as the chain rule, which we'll examine in more detail shortly. For our function, the derivative is f'(x) = -2e^{-2x}. The -2 factor appears from applying the chain rule, reflecting the impact of the coefficient in the exponent on the rate of change of the function.

Slope of a Tangent Line

The slope of a tangent line to a curve at a particular point represents the instantaneous rate of change of the function at that point. It equals the value of the derivative of the function at the point of tangency. Intuitively, this slope tells us how steep the curve is at that very point.

For the given function, f(x) = e^{-2x}, we calculate the slope of the tangent line at x = 0 by evaluating the derivative at this point. Substituting x = 0 into f'(x), we find that the slope, m, is -2. This means that at the point (0,1), the linear approximation of our curve is a straight line that falls two units vertically for every one unit it moves horizontally.

Chain Rule

The chain rule is a fundamental technique in calculus used to find the derivative of composite functions. In simple terms, it helps us differentiate functions that are made up of other functions, commonly referred to as 'function of a function'.

To apply the chain rule, we identify the 'outer' and 'inner' functions, differentiate each of them separately, and then multiply their derivatives together. In our case, the outer function is e^u (with u = -2x being the inner function). The derivative of e^u with respect to u is simply e^u, and the derivative of u with respect to x is -2. Hence, the derivative of f(x), using the chain rule, is -2e^{-2x}, the product of the derivative of e^u and u with respect to x.

Point-Slope Form

The point-slope form is an incredibly useful tool for writing the equation of a line when you have a point on that line and its slope. The general form is y - y1 = m(x - x1), where (x1, y1) is the point on the line, and m is the slope of the line.

In our exercise, we have the point (0,1) and the slope m = -2. By substituting these values into the point-slope equation, we quickly write down the equation of the tangent line. However, as x1 = 0, this simplifies our equation to y - 1 = -2(x - 0), which leads us to the final form y = -2x + 1. This equation represents the line that just 'touches' the curve at the point (0,1) and has the same instantaneous rate of change as the curve at that point, meaning it's tangent to the curve.