91Ó°ÊÓ

Understanding the derivative of a function is crucial in calculus, especially when dealing with tangent lines. In the context of our problem, the derivative of the given function \( f(x) = \frac{1}{x} \), represents the slope of the tangent line at any point on the graph of \( f \).

To calculate the derivative of \( f(x) = \frac{1}{x} \) or \( x^{-1} \) as written in its power form, we apply the power rule, which states that \( f'(x) = nx^{n-1} \). This results in \( f'(x) = -x^{-2} \) or \( -\frac{1}{x^2} \), which will be used to find the slope of the tangent line at any point \( x_0 \).

The process of derivative calculation is a fundamental concept in calculus and sets the stage for solving more complex problems like finding the equation of a tangent line to a curve at a given point.

The point-slope form is an equation that gives a straight line's slope when a point on the line is known. In algebraic terms, the point-slope form can be expressed as \( y - y_1 = m(x - x_1) \), where \( m \) represents the slope of the line, and \( (x_1, y_1) \) is the given point through which the line passes.

In our example, the slope \( m \) is given by the derivative \( f'(x_0) \) at the point \( P(x_0, y_0) \) on the graph. By substituting \( m \) with \( f'(x_0) \) and \( (x_1, y_1) \) with \( (x_0, y_0) \), we acquire the precise formula for the tangent line at \( P \) that passes through the external point \( Q \) given in the problem.

The point-slope form is a straightforward and powerful tool for creating linear equations, especially when we're focused on the geometric properties of the tangent line and its intersection with other points.

Solving for the equation of a tangent line to a function at a particular point involves using both the derivative of the function and the point-slope form of a line. As outlined in the provided exercise, once we have the derivative (representing the slope), and we identify a point on the curve, we can determine the equation of the tangent line easily.

After finding the derivative and writing the equation in point-slope form, we substitute in the fixed point \( Q(-2,4) \) to solve for \( x_0 \). When we solve for \( x_0 \), we determine where the tangent line to the curve will go through this designated point \( Q \). In our case, the point \( P(1,1) \) is found to be the point on \( f \) where the tangent passes through \( Q \) based on the given conditions.

This method is a key aspect of solving tangent lines in calculus, providing valuable insight into how curves behave locally and how they interact with other geometric entities.