Problem 14 - University Calculus: Early Transcendentals

Question

Find the slope of the function's graph at the given point. Then find an equation for the line tangent to the graph there. $$g(x)=\frac{8}{x^{2}}, \quad\quad(2,2)$$

Step-by-Step Solution

Answer

The slope at (2, 2) is -2, and the equation of the tangent line is \(y = -2x + 6\).

Step 1: Recognize the function and point

We have the function \( g(x) = \frac{8}{x^2} \) and we are given the point (2, 2). We will find the slope of the tangent line at this point and then form the equation of the tangent line.

Step 2: Compute the derivative

To find the slope of the tangent line, we first need the derivative of the function \( g(x) = \frac{8}{x^2} \). This can be rewritten as \( g(x) = 8x^{-2} \). Using the power rule for derivatives, compute \( g'(x) \):\[ g'(x) = -16x^{-3} = \frac{-16}{x^3}. \]

Step 3: Evaluate the derivative at the point

Next, evaluate the derivative at \( x = 2 \) to find the slope of the tangent line: \[ g'(2) = \frac{-16}{2^3} = \frac{-16}{8} = -2. \]So, the slope of the tangent line at the point (2, 2) is \(-2\).

Step 4: Use point-slope form to find the tangent line equation

Using the point-slope form of a line equation, \( y - y_1 = m(x - x_1) \), where \( m \) is the slope, and \( (x_1, y_1) \) is the point (2, 2), substitute the known values:\[ y - 2 = -2(x - 2). \]

Step 5: Simplify the equation

Simplify the equation from the previous step:\[ y - 2 = -2x + 4 \]\[ y = -2x + 6. \]This is the equation of the tangent line to the graph at the given point.

Key Concepts

Tangent Line

A tangent line is a straight line that touches a curve at a single point without crossing it. It can be thought of as the line that "just touches" the curve at a particular spot. The importance of tangent lines lies in their ability to represent the instantaneous rate of change of the function at a given point.

To find a tangent line, you need to determine two things:

• The point of tangency - where the line touches the curve, e.g., (2, 2) in our exercise.
• The slope of the tangent line - which is derived from the function's derivative at the specific point of tangency.

The equation of the tangent line can then be constructed using the point-slope form of a line, which requires both the slope and a point on the line.

Derivative of a Function

The derivative of a function measures how the function's output value changes as its input changes. It can be thought of as the "rate of change" or "slope" of the function at any given point. This is crucial in calculus because it provides insight into the function's behavior.

For a function like \( g(x) = \frac{8}{x^2} \), taking the derivative, \( g'(x) \), gives us a new function that tells us the slope of the tangent line at any point \( x \). The derivative is a central tool in calculus and is used to find instantaneous rates of change, optimize problems, and predict future behavior of functions.

In our specific case, we calculated \( g'(x) = \frac{-16}{x^3} \) which allowed us to determine the slope of the tangent line at \( x = 2 \).

Power Rule for Derivatives

The power rule is a basic formula in calculus used to find the derivative of functions of the form \( x^n \), where \( n \) is any real number. It states that the derivative is calculated by multiplying the power by the coefficient and then reducing the power by one. Mathematically, this is expressed as: \[\frac{d}{dx}(x^n) = n \cdot x^{n-1}.\]

This rule significantly simplifies the process of differentiation. For example, in transforming \( g(x) = 8x^{-2} \), applying the power rule gives: \[g'(x) = -2 \cdot 8x^{-3} = -16x^{-3} \text{ or } \frac{-16}{x^3}.\]

The power rule is fundamental because it provides a straightforward way to differentiate most polynomial expressions, saving time and reducing complexity when working with derivatives.