Problem 26 - Precalculus Mathematics for Calculus

Question

(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

(a) \( g'(a) = -\frac{2}{(2a-1)^2} \). (b) Tangent lines are \(y = -\frac{2}{9}x - \frac{11}{9}\), \(y = -2x + 1\), \(y = -\frac{2}{3}x + \frac{5}{3}\).

Step 1: Find the Derivative g'(x)

To find the derivative of \( g(x) = \frac{1}{2x-1} \), use the chain rule. The function can be rewritten as \( g(x) = (2x-1)^{-1} \). The derivative \( g'(x) \) is: \[ g'(x) = -1 \cdot (2x-1)^{-2} \cdot 2 = -\frac{2}{(2x-1)^2} \]

Step 2: Evaluate g'(a)

Substitute \(x = a\) into the derivative \(g'(x)\) found in Step 1. This gives us: \[ g'(a) = -\frac{2}{(2a-1)^2} \]

Step 3: Calculate Slope of Tangent Lines

For each \(x\)-coordinate, substitute the value into \(g'(x)\) to find the slope of the tangent line:- For \(x = -1\), \(g'(-1) = -\frac{2}{(-3)^2} = -\frac{2}{9}\).- For \(x = 0\), \(g'(0) = -\frac{2}{1} = -2\).- For \(x = 1\), \(g'(1) = -\frac{2}{3}\).

Step 4: Find Tangent Line Equations

Use the point-slope form of a line, \(y - y_1 = m(x - x_1)\), with \((x_1, y_1)\) being a point on the curve:- At \(x = -1\): The point \((-1, g(-1)) = (-1, -\frac{1}{3})\) and slope \(-\frac{2}{9}\). The line equation is \(y + \frac{1}{3} = -\frac{2}{9}(x + 1)\).- At \(x = 0\): The point \((0, g(0)) = (0, 1)\) and slope \(-2\). The line equation is \(y - 1 = -2x\).- At \(x = 1\): The point \((1, g(1)) = (1, \frac{1}{1})\) and slope \(-\frac{2}{3}\). The line equation is \(y - 1 = -\frac{2}{3}(x - 1)\).

Step 5: Graph g and Tangent Lines

Plot the function \(g(x) = \frac{1}{2x-1}\) using its general shape and asymptotes. For each tangent line, plot their respective equations, which are: - \(y = -\frac{2}{9}x - \frac{11}{9}\) for \(x = -1\).- \(y = -2x + 1\) for \(x = 0\).- \(y = -\frac{2}{3}x + \frac{5}{3}\) for \(x = 1\). This gives a visual of how the tangent lines touch the curve.

Key Concepts

Derivative

In calculus, the derivative of a function is a measure of how the function's value changes as its input changes. Essentially, it tells us the rate at which the function is changing. With the function provided, \(g(x) = \frac{1}{2x-1}\), taking the derivative involves applying the chain rule. By rewriting it in exponent form, we get \((2x-1)^{-1}\). The derivative, \(g'(x)\), is found using the formula:

• Differentiate the outer function \([-1\cdot(\text{whatever})^{-2}]\).
• Differentiate the inner function \(2x-1\), which gives \(2\).
• The chain rule tells us to multiply these derivatives. The result is \(-\frac{2}{(2x-1)^2}\).

Understanding derivatives is critical as they are used to approximate function values and understand their behavior.

Tangent Line

A tangent line is a straight line that touches a function's curve at exactly one point. This line reflects the instantaneous rate of change of the function at that point, which means its slope is the same as the derivative of the function at the same point. Once you know the slope from the derivative \(g'(x)\), creating the equation for the tangent line uses:

• A known point on the curve \((x_1, g(x_1))\).
• Apply the point-slope formula: \(y - y_1 = m(x - x_1)\).

For example, using the derivative calculated:- At \(x=-1\), the slope \(g'(-1) = -\frac{2}{9}\), giving the equation \(y + \frac{1}{3} = -\frac{2}{9}(x + 1)\).- At \(x=0\), the slope \(g'(0) = -2\), giving a simpler version: \(y - 1 = -2x\).- At \(x=1\), the slope \(g'(1) = -\frac{2}{3}\), with \(y - 1 = -\frac{2}{3}(x - 1)\).

Graphing Functions

Graphing functions like \(g(x) = \frac{1}{2x-1}\) helps to visualize the behavior of the function. This particular function is a rational function, characterized by a vertical asymptote where the denominator equals zero (here, \(x = \frac{1}{2}\)). To graph it:

• Identify asymptotes and intercepts to guide your plot.
• Draw a smooth curve that approaches but never crosses the asymptotes.

Adding tangent lines to the graph involves:

• Drawing each line using its equation.
• Ensuring each line touches the curve at the specific point of tangency.

This provides an illustration of how the tangent lines reflect changes in the graph's direction at their specific points.

Chain Rule

The chain rule is a fundamental tool in calculus used to differentiate composite functions. When a function is composed of two or more functions, the chain rule provides a straightforward method to differentiate it. For the function \(g(x) = \frac{1}{2x-1}\), which can also be expressed as \((2x-1)^{-1}\), the chain rule applies. Here’s how it works:

• Identify the "outer" function \((u^{-1})\) and the "inner" function \((u = 2x-1)\).
• The derivative of the outer function \((u^{-1})\) is \(-u^{-2}\).
• Then find the derivative of the inner function, \(2x-1\), which is \(2\).
• Multiply the derivatives of the outer and inner functions: \(-1 \cdot u^{-2} \cdot 2 = -\frac{2}{(2x-1)^2}\).

By applying the chain rule, we get a precise calculation of how changes in \(x\) affect the function. Mastering this rule is essential for tackling more complex differentiation problems.