Problem 32 - Precalculus: Mathematics for Calculus

Question

Tangent Lines (a) If \(g(x)=1 /(2 x-1),\) find \(g^{\prime}(a)\) (b) Find equations of the tangent lines to the graph of \(g\) at the points whose \(x\) -coordinates are \(-1,0,\) and 1 (c) Graph \(g\) and the three tangent lines.

Step-by-Step Solution

Answer

The derivative is \( g'(a) = \frac{-2}{(2a-1)^2} \). Tangent lines are \( y = -\frac{2}{9}x - \frac{5}{9} \), \( y = -2x - 1 \), and \( y = -2x + 3 \). Graph accordingly.

Step 1: Find the Derivative g'(x)

To find the equation of the tangent line to the graph of a function, the first step is to determine the derivative of the function. Start by noting the given function: \( g(x) = \frac{1}{2x-1} \). Using the quotient rule for derivatives, if \( h(x) = \frac{f(x)}{u(x)} \), then: \[ \frac{d}{dx} \left( \frac{f(x)}{u(x)} \right) = \frac{f'(x)u(x) - f(x)u'(x)}{[u(x)]^2} \]Here, \( f(x) = 1 \) and \( u(x) = 2x-1 \). Hence, \( f'(x) = 0 \) and \( u'(x) = 2 \).Substitute these into the formula:\[ g'(x) = \frac{0 \cdot (2x-1) - 1 \cdot 2}{(2x-1)^2} = \frac{-2}{(2x-1)^2} \] Thus, \( g'(x) = \frac{-2}{(2x-1)^2} \).

Step 2: Evaluate g'(a)

Next, for part (a), we evaluate \( g'(a) \) by substituting \( a \) into the derivative:\[ g'(a) = \frac{-2}{(2a-1)^2} \]This expression represents the rate of change or the slope of the tangent line at \( x = a \).

Step 3: Find Tangent Line Equations

For part (b), we need to find the equations of the tangent lines to the graph of \( g(x) \) at points \( x = -1, 0, \) and \( 1 \). - **At \( x = -1 \):** - \( g(-1) = \frac{1}{-3} = -\frac{1}{3} \) - Slope, \( g'(-1) = \frac{-2}{(-3)^2} = -\frac{2}{9} \) - Equation: \( y + \frac{1}{3} = -\frac{2}{9}(x + 1) \) - Simplified: \( y = -\frac{2}{9}x - \frac{5}{9} \)- **At \( x = 0 \):** - \( g(0) = -1 \) - Slope, \( g'(0) = \frac{-2}{1} = -2 \) - Equation: \( y + 1 = -2(x - 0) \) - Simplified: \( y = -2x - 1 \)- **At \( x = 1 \):** - \( g(1) = 1 \) - Slope, \( g'(1) = \frac{-2}{1} = -2 \) - Equation: \( y - 1 = -2(x - 1) \) - Simplified: \( y = -2x + 3 \)

Step 4: Graph the Function and Tangents

Finally, for part (c), graph the function \( g(x) = \frac{1}{2x-1} \) along with the tangent lines.The main features of \( g(x) \) include a vertical asymptote at \( x = \frac{1}{2} \) and horizontal asymptote at \( y = 0 \). Mark the points \((-1, -\frac{1}{3})\), \( (0, -1) \), and \( (1, 1) \) where the tangent lines intersect the graph, and draw lines with slopes \( -\frac{2}{9} \), \(-2\), and \(-2\), respectively.

Key Concepts

Derivative

Understanding derivatives is key in calculus, especially when dealing with tangent lines. The derivative of a function gives us the slope of the tangent line at any given point on the function's graph. For the function \( g(x) = \frac{1}{2x-1} \), its derivative \( g'(x) \) is calculated using the rules of differentiation. Namely, when you find the rate at which \( g(x) \) changes at each \( x \), you discover the slope of the line that just "touches" the function at that point without cutting across it. In this context, the derivative \( g'(x) \) helps us compute the slope of the tangent line, enabling us to write the equation of the tangent line in point-slope form: \( y - y_1 = m(x - x_1) \), where \( m \) is the slope (or the derivative value) and \( (x_1, y_1) \) is the point of tangency on the graph of the function.

Quotient Rule

The quotient rule is an essential tool in differentiation when dealing with functions that are fractions—ratios of two functions. If a function is expressed as \( h(x) = \frac{f(x)}{u(x)} \), the quotient rule helps in finding its derivative, which is given by:

• \( \frac{d}{dx} \left( \frac{f(x)}{u(x)} \right) = \frac{f'(x)u(x) - f(x)u'(x)}{[u(x)]^2} \)

This formula may look complex, but breaking it down step by step can make it simpler. Find the derivatives of the numerator and the denominator separately, then plug them into this formula to find the derivative of the entire function. In our function, \( g(x) = \frac{1}{2x-1} \), applying the quotient rule involves realizing the numerator is a constant (1) and handling the denominator's derivative carefully to find \( g'(x) = \frac{-2}{(2x-1)^2} \). Each piece is essential to determine accurately the nature of the tangent at any given point on the graph of \( g \).

Graphing Functions

Graphing a function helps visualize how it behaves and where the tangent lines interact with it. For the function \( g(x) = \frac{1}{2x-1} \), we derive some properties that contribute to its graph: a vertical asymptote at \( x = \frac{1}{2} \) and a horizontal asymptote at \( y = 0 \). Understanding where these asymptotes are is crucial because they demonstrate how the function behaves as \( x \) approaches these values. Additionally, by calculating points where tangent lines intersect the graph, such as \((-1, -\frac{1}{3})\), \( (0, -1) \), and \( (1, 1) \), we assure the quality of the tangent lines fittingly representing the rate of change at those points.Furthermore, these tangents, with respective slopes, uniquely describe the steepness or flatness of \( g(x) \) as it passes through important points.

Calculus

Calculus is the broader field that encompasses all these individual concepts and operations, like derivatives and tangent lines. It provides us with the framework to analyze and understand changes within continuous functions. The branch of calculus used here is differential calculus, which focuses on understanding how functions change over intervals — that's where tangent lines, slopes, and rates of change originate. Exploring calculus allows mathematicians and students to explore more complex scenarios like optimizing real-world systems, understanding natural phenomena, and delving into the intricacies of motion and change. By mastering the foundational procedures like finding derivatives and applying tools such as the quotient rule, students can apply these skills to both theoretical and practical problems.