Problem 22 - College Algebra and Calculus: An Applied Approach

Question

In Exercises, determine an equation of the tangent line to the function at the given point. $$ y=\left(e^{4 x}-2\right)^{2}, \quad(0,1) $$

Step-by-Step Solution

Answer

The equation of the tangent line to the function at the point (0,1) is \(y = -8x + 1.\)

Step 1: Calculate Derivative

Begin by calculating the derivative of \(y = (e^{4x} - 2)^2\). This will require the Chain Rule. The outer function is \(f(u) = u^2\) and the inner function is \(u = e^{4x} - 2\). By applying Chain Rule, we get \[y' = 2(e^{4x} - 2)(4e^{4x}) = 8e^{4x}(e^{4x} - 2).\]

Step 2: Evaluate the derivative at the given point

For the given point \((0,1)\), the x-coordinate is 0. Plug this value into the derivative to find slope of the tangent line at this point: \[m = 8e^{4*0}(e^{4*0} - 2) = 8e^{0}(e^{0} - 2) = 8(1)(1-2) = -8.\] So the slope of the tangent line at the point \((0,1)\) is -8.

Step 3: Find the equation of the tangent line

The equation of a line is given by \(y = mx + b\) where m is the slope and b is the y-intercept. We already know the slope (m = -8) from the previous step. Since the line passes through the given point \((0,1)\), when x = 0, y = 1. So we get: \[1 = -8*0 + b,\] solving for b, we find b = 1. Therefore, the equation of the tangent line to the function at the given point is \(y = -8x + 1.\)

Key Concepts

Chain Rule

Understanding the Chain Rule is crucial for differentiating compositions of functions, as often found in calculus. Imagine you have a series of connected gears, with each influencing how the next one turns. Similarly, the Chain Rule handles functions working inside other functions. In the given exercise, where we have a function \(y = (e^{4x} - 2)^2\), we notice it's a composite function: an exponential function inside a square function.

To differentiate this using the Chain Rule, identify the 'outer function' (the square function, \(f(u) = u^2\)) and the 'inner function' (the exponential function, \(u = e^{4x} - 2\)) separately. Then, take the derivative of the outer function with respect to the inner one, \(f'(u)\), and multiply it by the derivative of the inner function with respect to \(x\), \(u'(x)\). For our exercise, \(f'(u) = 2u\) and \(u'(x) = 4e^{4x}\), which upon applying the Chain Rule gives us \(y' = 2(e^{4x} - 2) \cdot 4e^{4x}\).

Derivative

The derivative represents the rate at which a function's output is changing at any given point, effectively giving us the slope of the tangent line at that point. In simpler terms, it's like the speedometer of your car indicating how fast you're going at a single moment in time.

In the context of the problem, calculating the derivative of \(y = (e^{4x} - 2)^2\) tells us how quickly \(y\) changes as \(x\) changes. It's a fundamental tool in calculus because it provides information about the function's instantaneous rate of change, which is essential for finding slopes of tangent lines, determining function behavior, and solving practical problems involving rates.

Slope of the Tangent Line

The slope of the tangent line to a curve at a particular point represents the steepness or incline of the line just touching the curve at that point. This slope is a snapshot of the curve's inclination at the exact location and is given by the value of the function's derivative at that point.

From our exercise, after finding the derivative \(y'\), we needed to determine the specific slope of the tangent line at the point \(0, 1\). Assimilating the slope to a real-world concept, it's like figuring out the precise angle at which a hill starts at a specific point. The derivative provided a general formula for the slope, and by using the coordinates of the point \(0, 1\), we were able to calculate that the slope is -8, meaning the tangent line is descending sharply.

Function Derivative Evaluation

Finding the Slope at a Specific Point

Once you have the derivative, it needs to be evaluated at a specific point to find the slope of the tangent line there. This is done by substituting the x-coordinate of the point into the derivative formula.

For example, in our given problem, we evaluated the derivative at the point \(x = 0\) and found the slope to be -8. This process is akin to taking a general rule and applying it to a specific, real-world instance to get a precise answer.

Determining the Equation of a Line

With the slope known, you can then use the point-slope form of a line's equation to find the actual equation of the tangent line. The problem required us to find the equation passing through the given point with the slope we calculated, ultimately leading us to \(y = -8x + 1\). This equation gives a linear relationship that locally approximates the behavior of the function around the point \(0, 1\).