Problem 60 - Calculus and its applications

Question

Find an equation of the tangent line to the graph of \(f(x)=\frac{1}{x^{2}}\) a) at (1,1)\(;\) b) at \(\left(3, \frac{1}{9}\right)\) c) at \(\left(-2, \frac{1}{4}\right)\).

Step-by-Step Solution

Answer

a) \(y = -2x + 3\); b) \(y = -\frac{2}{27}x + \frac{1}{3}\); c) \(y = \frac{1}{4}x + \frac{1}{2}\)."

Step 1: Understand the Problem

We need to find the tangent line to the function \( f(x) = \frac{1}{x^2} \) at the given points. The tangent line will have the same slope as the derivative of the function at that point.

Step 2: Find the Derivative

First, let's compute the derivative of \( f(x) = \frac{1}{x^2} \). We can rewrite \( f(x) \) as \( f(x) = x^{-2} \). The derivative, \( f'(x) \), using the power rule, is \( -2x^{-3} \) or equivalently \( f'(x) = -\frac{2}{x^3} \).

Step 3: Calculate the Slope at Each Point

To find the slope of the tangent line at a given point, substitute the x-value into the derivative:- For point (1,1), the slope \( m = f'(1) = -\frac{2}{1^3} = -2 \).- For point \( (3, \frac{1}{9}) \), the slope \( m = f'(3) = -\frac{2}{3^3} = -\frac{2}{27} \).- For point \( (-2, \frac{1}{4}) \), the slope \( m = f'(-2) = -\frac{2}{(-2)^3} = \frac{1}{4} \).

Step 4: Use the Point-Slope Form to Find the Tangent Lines

The point-slope form of the equation of a line is \( y - y_1 = m(x - x_1) \), where \( m \) is the slope, and \( (x_1, y_1) \) is the given point.- For point (1,1), substitute \( m = -2 \), \( x_1 = 1 \), \( y_1 = 1 \): \[ y - 1 = -2(x - 1) \].- For point \( (3, \frac{1}{9}) \), substitute \( m = -\frac{2}{27} \), \( x_1 = 3 \), \( y_1 = \frac{1}{9} \): \[ y - \frac{1}{9} = -\frac{2}{27}(x - 3) \].- For point \( (-2, \frac{1}{4}) \), substitute \( m = \frac{1}{4} \), \( x_1 = -2 \), \( y_1 = \frac{1}{4} \): \[ y - \frac{1}{4} = \frac{1}{4}(x + 2) \].

Step 5: Solve for y to Find Explicit Form of Tangent Lines

Rearrange each equation to the form \( y = mx + b \):- For point (1,1): \[ y = -2x + 3 \].- For point \( (3, \frac{1}{9}) \): \[ y = -\frac{2}{27}x + \frac{1}{9} + \frac{6}{27} = -\frac{2}{27}x + \frac{9}{27} = -\frac{2}{27}x + \frac{1}{3} \].- For point \( (-2, \frac{1}{4}) \): \[ y = \frac{1}{4}x + \frac{1}{2} \].

Key Concepts

Derivative

The concept of a derivative is crucial when trying to find a tangent line. Essentially, the derivative gives us the slope of the tangent line to a function at any point. Imagine a curve on a graph; the tangent line is a straight line that just "touches" the curve at a single point without crossing it.
To determine this line's slope, we need the derivative of the function. For a function like \( f(x) = \frac{1}{x^2} \), the derivative tells us how \( y \) changes with \( x \).
By rewriting the function as \( f(x) = x^{-2} \), we make it easier to apply the power rule, a common technique for finding derivatives. Understanding derivatives allows you to not only find the slope at specific points but also to understand how a function behaves overall.

Power Rule

The power rule is a straightforward method used in calculus to find the derivative of a function with a power of \( x \). It's expressed as: if \( f(x) = x^n \), then the derivative \( f'(x) = nx^{n-1} \).
In our case, \( f(x) = x^{-2} \). By applying the power rule, we differentiate to get \( f'(x) = (-2)x^{-3} \) or \( -\frac{2}{x^3} \). This gives us the derivative function, which can be applied to find slopes at any point on the curve.
For any function with a variable exponent, remembering the power rule allows you to quickly find the rate at which the function is changing at any given point.

Point-Slope Form

Once the slope of the tangent line is known, the point-slope form comes into play to write the equation of the line. The point-slope form equation is \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is a specific point on the line, and \( m \) is the slope.
Given a point, such as \( (1, 1) \), and a slope, such as \( -2 \), this form makes it simple to write the line's equation as \( y - 1 = -2(x - 1) \).
This step provides a straightforward method to transition from a calculated slope to a tangible equation representing the line.

Slope Calculation

Slope calculation is a significant step in finding the tangent line. Once you have the derivative, plug in the \( x \)-value of the point of interest to find the slope at that point. For the function \( f(x) = \frac{1}{x^2} \), the derivative \( f'(x) = -\frac{2}{x^3} \) helps us find the slopes at given points:

• At (1,1), the slope is \( -2 \).
• At \( (3, \frac{1}{9}) \), it's \( -\frac{2}{27} \).
• At \( (-2, \frac{1}{4}) \), it becomes \( \frac{1}{4} \).

Each calculation involves a substitution into the derivative and simplifies to determine how steep or flat the tangent at that specific point is. Understanding how to perform slope calculations ensures you have the correct, context-specific slope for further calculations like the equation of the tangent line.