Problem 20 - College Algebra and Calculus: An Applied Approach

Question

In Exercises, determine an equation of the tangent line to the function at the given point. $$ y=\frac{x}{e^{2 x}}, \quad\left(1, \frac{1}{e^{2}}\right) $$

Step-by-Step Solution

Answer

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

Step 1: Compute the Derivative

The derivative of the function \(y=\frac{x}{e^{2x}}\) can be found using the quotient rule, which states that the derivative of \(\frac{u}{v}\) is \(\frac{vu'-uv'}{v^2}\), where \(u',v'\) are the derivatives of \(u\) and \(v\) respectively. Here, \(u=x\), \(u'=1\), \(v=e^{2x}\), and \(v'=2e^{2x}\). Therefore, \(y'=\frac{e^{2x}*1 - x*2e^{2x}}{{(e^{2x})^2}} = \frac{e^{2x}(1 - 2x)}{e^{4x}} = \frac{1 - 2x}{e^{2x}}\).

Step 2: Find the Slope of the Tangent Line

The slope of the tangent at the point \((1, \frac{1}{e^{2}})\) can be found by substituting \(x=1\) into the derivative. Thus, the slope is \(y'(1)= \frac{1 - 2*1}{e^{2*1}} =\frac{1 - 2}{e^{2}} = -\frac{1}{e^{2}}\)

Step 3: Find the Equation of the Line

The equation of the line can be found using the point-slope form \(y - y1 = m(x - x1)\), where \((x1, y1)\) are the coordinates of the given point and \(m\) is the slope of the line. Therefore, the equation of the tangent is \(y - \frac{1}{e^{2}} = -\frac{1}{e^{2}}(x - 1)\)

Key Concepts

Derivative

The derivative gives us a way to understand how a function changes as the input changes. In simpler terms, it describes the slope of the function at any given point.
When dealing with functions, particularly in calculus, the derivative offers crucial insights into rates of change and the behavior of graphs. Calculating the derivative helps us find the slope of the tangent line at any point on the function, which is essential in determining the function's behavior.
For instance, in the example exercise, we work with the function \( y = \frac{x}{e^{2x}} \). To find its derivative, we use the rules of differentiation, such as the quotient rule, to grasp how the function behaves near \( x = 1 \). By understanding the derivative, we can predict how small changes in \( x \) affect \( y \). This understanding is powerful in applications ranging from physics to economics.

Quotient Rule

The quotient rule is a method in calculus for finding the derivative of a fraction, specifically when one function is divided by another.
According to the rule, if you have two functions \( u(x) \) and \( v(x) \), their quotient \( \frac{u}{v} \) has a derivative \( \frac{vu' - uv'}{v^2} \). This formula helps in cases where a simple power rule or product rule would not suffice.
In the example, to differentiate \( y = \frac{x}{e^{2x}} \), we identify \( u = x \) and \( v = e^{2x} \). We then compute their derivatives: \( u' = 1 \) and \( v' = 2e^{2x} \).
Plugging into the formula gives:

• Numerator: \( e^{2x} \cdot 1 - x \cdot 2e^{2x} = e^{2x} - 2xe^{2x} \)
• Denominator: \( (e^{2x})^2 = e^{4x} \)

So the derivative simplifies to \( \frac{1 - 2x}{e^{2x}} \), a crucial part of finding the tangent line's slope.

Point-Slope Form

The point-slope form is a tool used to write the equation of a line when a point on the line and the slope are known.
It's expressed as \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is a given point on the line, and \( m \) is the slope.
In our example, after calculating the slope at \( x = 1 \) as \( -\frac{1}{e^{2}} \), we can write the tangent line's equation using the point \( (1, \frac{1}{e^{2}}) \):

• Substitute: \( y - \frac{1}{e^{2}} = -\frac{1}{e^{2}}(x - 1) \)

This form is helpful because it directly incorporates the point and the slope, making it straightforward to visualize the line's steepness and direction.
Point-slope form is especially useful when dealing with tangents in calculus or linear equations in algebra.

Function Analysis

Function analysis involves examining a function's behavior and its characteristics, such as continuity, limits, and derivatives.
It's about understanding how the function behaves over different intervals and at specific points.
When analyzing the function \( y = \frac{x}{e^{2x}} \) at the point \( (1, \frac{1}{e^{2}}) \), steps include:

• Identifying critical points where the derivative is zero or undefined.
• Determining slopes and curves using the derivative.
• Checking for asymptotic behaviors as \( x \) approaches certain values.

Function analysis helps us ascertain where the function increases or decreases and where it may have maximum or minimum values.
For functions like \( \frac{x}{e^{2x}} \), understanding its growth can be tricky due to exponential terms, but derivative and tangent analysis give insight into its behavior around critical points like \( x = 1 \). This understanding is foundational to mastering calculus and applying it to real-world problems.