Problem 69 - College Algebra and Calculus: An Applied Approach

Question

In Exercises, find the slope of the graph at the indicated point. Then write an equation of the tangent line to the graph of the function at the given point. $$ f(x)=\ln \frac{5(x+2)}{x}, \quad(-2.5,0) $$

Step-by-Step Solution

Answer

The slope of the tangent line at the given point is 0 and the equation of the tangent line is y = 0.

Step 1: Find the derivative

Using the quotient rule, which states that \((f/g)' = (g*df/dx - f*dg/dx)/(g^2)\), the derivative of \(f(x)=\ln \frac{5(x+2)}{x}\) is \[f'(x)=\frac{x*(5)+5(x+2)*1}{x*\frac{5(x+2)}{x}} - \frac{\frac{5(x+2)}{x}*1}{x^2}\]. Using the chain rule and simplification eventually gives us \(f'(x) = \frac{_5/(x+2) - \ln{(5(x+2)/x}}{x}\)

Step 2: Find the slope at the given point

Substitute your x-value, which is -2.5 from the point (-2.5,0), into the derivative to find the slope of the tangent line at that point. This gives you \(f'(-2.5) = \frac{5/(-2.5+2) - \ln{5(-2.5+2)/(-2.5)}}{-2.5} = 0 \)

Step 3: Write the equation of the tangent line

With the knowledge that the slope at the point is 0, and the coordinates of the point are (-2.5,0), you can write the equation of the tangent line. Using the point slope form, \(y - y_1 = m(x - x_1)\), the equation of the tangent line is \(y - 0 = 0*(x+2.5)\) which simplifies to \(y = 0\)

Key Concepts

Slope

In mathematics, the slope of a line on a graph represents how steep the line is. It is calculated as the ratio of the change in the vertical direction (the rise) to the change in the horizontal direction (the run). The slope is computed using the formula:

• Slope, or m, is given by \[m = \frac{\text{rise}}{\text{run}} = \frac{\Delta y}{\Delta x}= \frac{y_2 - y_1}{x_2 - x_1}\]

A positive slope means the line is going upwards as you move from left to right, while a negative slope indicates the line descends. If the slope is zero, the line is perfectly horizontal. When finding the slope of a tangent line at a specific point on a curve, you need to determine the derivative first. Then, plug in the x-value of the point to find the slope at that precise spot on the graph.

Derivative

The derivative of a function is a fundamental concept in calculus that measures how a function's output changes as its input changes. In simpler terms, it provides the rate of change or the slope of the function at any point. When you differentiate a function, you essentially calculate the slope of the tangent line for every point on the curve.

• Notation for derivatives includes \(f'(x)\) or \(\frac{dy}{dx}\)

To find a derivative, we use rules like the power rule, product rule, quotient rule, and chain rule. In this exercise, we used the quotient rule due to the nature of the function being a quotient, which requires specific steps to find the derivative correctly.

Quotient Rule

The quotient rule is a technique for finding the derivative of a function that is the ratio of two differentiable functions. It's essential when you have one function being divided by another. The rule is expressed as:\[\left(\frac{f}{g}\right)' = \frac{g \cdot f' - f \cdot g'}{g^2}\]where \(f\) and \(g\) are functions of \(x\), and \(f'\) and \(g'\) are their respective derivatives. In this exercise, we applied the quotient rule to differentiate the function \(f(x) = \ln \frac{5(x+2)}{x}\). This involved calculating the derivatives of both the numerator and the denominator and then substituting them into the quotient rule formula. Using the quotient rule can sometimes be complex, especially if the functions involved also require the use of other rules like the chain rule.

Chain Rule

The chain rule is a critical tool in calculus used for differentiating composite functions. A composite function is one that is made up of two or more functions. The chain rule helps find the derivative of such functions by "chaining" the derivatives of the inner and outer functions together.The chain rule is expressed as:\[\text{If } y=f(u) \text{ and } u=g(x), \text{ then } \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\]To utilize the chain rule, you first differentiate the outer function with respect to the inner function, and then multiply by the derivative of the inner function with respect to \(x\). In the given solution, the chain rule was used after applying the quotient rule, since the expression involved a composite function due to the natural logarithm. This ensures that each part of the function is correctly differentiated and contributes to finding the slope of the tangent at a specific point.