Problem 18 - College Algebra and Calculus: An Applied Approach

Question

In Exercises, determine an equation of the tangent line to the function at the given point. $$ g(x)=e^{x^{3}}, \quad\left(-1, \frac{1}{e}\right) $$

Step-by-Step Solution

Answer

The equation of the tangent line to the function \(g(x)=e^{x^{3}}\) at the point \((-1, 1/e)\) is \(y = 3x/e + 4/e\).

Step 1: Find the Derivative of the Given Function

The first step is to find the derivative of the function \(g(x)=e^{x^{3}}\). Using chain rule, the derivative of \(g(x)\) denoted \(g'(x)\) is: \(g'(x) = e^{x^{3}} * 3x^{2}\)

Step 2: Calculate the Slope at the Given Point

The derivative at a particular point gives the slope of the tangent at that point. Substituting \(x=-1\) into \(g'(x)\): \(g'(-1) = e^{-1^{3}} * 3(-1)^{2} = e^{-1} * 3 = 3/e\)

Step 3: Find Equation of the Tangent Line

The equation of the tangent line is given by the slope-point form: \(y - y_{1} = m(x - x_{1})\) where \(m\) is the slope of the line and \((x_{1}, y_{1})\) is the given point. Substituting \(m=3/e\), \(x_{1}=-1\), and \(y_{1}=1/e\) gives: \(y - 1/e = 3/e * (x - (-1))\). Simplifying this gives the equation of the tangent line: \(y = 3x/e + 4/e\)

Key Concepts

Derivative Calculation

Understanding the concept of derivative calculation is fundamental in calculus. It tells us how a function changes at any given point. For our example,
the function is given by \(g(x)=e^{x^{3}}\). To calculate its derivative, we involve the rate of change of e raised to a power. The derivative of \(e^x\) with respect to x is simply \(e^x\), but since we have a more complex exponent \(x^3\), we must use the chain rule.

The chain rule is a technique for differentiating composite functions, and as we apply it to our given function, we calculate the derivative of the outer function and multiply it by the derivative of the inner function, \(x^3\). The result is \(g'(x) = e^{x^{3}} \cdot 3x^{2}\), which simplifies the process of finding the instantaneous rate of change of \(g(x)\) at any point.

Chain Rule

The chain rule is a powerful tool in differentiation. This rule is used when dealing with the derivative of a function composed of two or more functions.
The idea is to take the derivative of the outer function (let's call it 'f') with respect to the inner function (we'll call this one 'g'), and then multiply it by the derivative of 'g' with respect to 'x'. Written generally:
\(\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\).

For our exercise, f corresponds to \(e^u\), and g is our \(x^{3}\). So, according to the chain rule, the derivative \(g'(x)\) is the product of the derivative of f with respect to u, and the derivative of g with respect to x, leading to \(e^{x^{3}} * 3x^{2}\). This calculation thus allows us to find how quickly our function is changing at any x-value.

Slope of Tangent

The slope of the tangent line to a curve at a particular point indicates how steep the line is at that exact location.
In our function, finding the slope means evaluating the derivative at the given point. Recall that the derivative represents the slope of the tangent line at any point of the function. If we label the point as \(x_{1}\), the slope \(m\) is given by \(g'(x_{1})\). After calculating the derivative as demonstrated in previous sections, we substitute the given x-coordinate of our point, \(-1\), into the derivative to find the specific slope.

So, \(g'(-1) = e^{-1^{3}} * 3(-1)^{2} = \frac{3}{e}\), resulting in a slope of \(\frac{3}{e}\). With this value, we now have an essential part of the information required to write the equation for the tangent line.

Point-Slope Form

The point-slope form of the equation of a line is perhaps the most direct method to writing the equation of a line given a point and a slope. The formula is as follows:
\(y - y_{1} = m(x - x_{1})\).

Here, \(m\) represents the slope of the line, and \((x_{1}, y_{1})\) denotes a point on the line. Using the slope \(\frac{3}{e}\) from our tangent slope calculation and the coordinates of our given point \((-1, \frac{1}{e})\), we substitute these values into the point-slope formula.

Rearranging and simplifying, we get the line's equation: \(y = \frac{3x}{e} + \frac{4}{e}\). Thus, we've not only found the slope but have constructed the exact line that just graces the curve at our specified point, fully completing the picture of the tangent.