Problem 141 - Calculus Volume 1

Question

Find the equation of the line passing through the point \(P(3,3)\) and tangent to the graph of \(f(x)=\frac{6}{x-1}\)

Step-by-Step Solution

Answer

Solve the equation for \( a \), find the slope, and write the tangent line's equation.

Step 1: Understand the Problem

We are asked to find the equation of a line that is tangent to the graph of the function \( f(x) = \frac{6}{x-1} \) and passes through the point \( P(3,3) \). This means that the line will intersect the curve at exactly one point.

Step 2: Find the Derivative of the Function

To find the equation of the tangent line, we need the slope at the tangent point. First, find the derivative of \( f(x) = \frac{6}{x-1} \). The derivative \( f'(x) \) is computed using the quotient rule: \[ f'(x) = \frac{-6}{(x-1)^2} \]

Step 3: Set the Equation of the Tangent Line

The equation of the tangent line in point-slope form is \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \((x_1, y_1)\) is the point on the line. We will determine \( m \) and \( (x_1, y_1) \) so that the line passes through (3, 3) and is tangent to the curve.

Step 4: Find a Function Value and Slope at the Tangent Point

Suppose the tangent point is \( Q(a, f(a)) \) on the function. The slope \( m \) is given by \( f'(a) \). We set up the condition that our line, passing through \( (3, 3) \) with slope \( f'(a) \), is tangent to \( f(x) \):\( 3 - \frac{6}{a-1} = \frac{-6}{(a-1)^2} (3-a) \).

Step 5: Solve the Equation for ‘a’

By solving the equation from Step 4, you find \( a \). Simplifying gives:\( 3(a-1)^2 - 6 = -6(3-a)(a-1) \)Expanding and solving can be simplified further to find values for \( a \) that make the equation true.

Step 6: Find Corresponding y-value and Slope

Once \( a \) is found, compute \( f(a) \) to get \( y_1 \). Use \( a \) to find slope \( m = f'(a) \). Remember we already have the point \( y_1 = \frac{6}{a-1} \).

Step 7: Write the Equation of the Tangent Line

Substitute \( m \) and point \((3, 3)\) into the equation of the line \( y - 3 = m(x - 3) \). After calculating the slope, write the equation of the tangent line. Simplify to get it in standard or slope-intercept form if needed.

Key Concepts

Understanding Derivatives

Derivatives are a fundamental concept in calculus. They represent the rate at which a function is changing at any given point. For tangent lines, the derivative at a point gives us the slope of the tangent line to the curve at that point. The derivative is essential when we want to know how steep the graph of a function is at a particular location. In our exercise, we need to find the derivative of the function \(f(x) = \frac{6}{x-1}\) to determine the slope of the tangent line. One way to find this derivative is by applying the quotient rule since the function is a fraction (or quotient) of two expressions.

Point-Slope Form Explained

Point-slope form is a way of expressing the equation of a straight line. It is particularly useful when we know one point on the line and the slope of the line. The general equation is \(y - y_1 = m(x - x_1)\), where \(m\) is the slope of the line and \((x_1, y_1)\) is a point on the line.In our task, once we identify the slope \(m\) from the derivative of the function, we use the point \((3, 3)\), which the tangent line must pass through, to form the equation. This method allows us to directly plug in values to find the line's equation efficiently.This form is preferred in calculative problems since it simplifies the process of writing the equation when the slope and a single point are given.

Applying the Quotient Rule

The quotient rule is used to find the derivative of a function that is the division of two functions. It offers a structured approach for correctly differentiating such expressions. In a mathematical form, if you have a function \( h(x) = \frac{u(x)}{v(x)} \), the derivative \( h'(x) \) is given by:\[ h'(x) = \frac{u'(x) v(x) - u(x) v'(x)}{[v(x)]^2}\]For our specific function \( f(x) = \frac{6}{x-1} \), apply the quotient rule: - Numerator \(u(x) = 6\) has a derivative \(u'(x) = 0\) since it's constant. - Denominator \(v(x) = x-1\) has a derivative \(v'(x) = 1\).Substituting these into the quotient rule provides \(f'(x) = \frac{0\cdot(x-1) - 6\cdot1}{(x-1)^2} = \frac{-6}{(x-1)^2}\). This derivative tells us how the function \( f(x) \) is changing at any point \(x\), and is crucial for finding the equation of the tangent line at the desired tangency point.