Problem 13 - University Calculus: Early Transcendentals

Question

Find the slope of the function's graph at the given point. Then find an equation for the line tangent to the graph there. $$g(x)=\frac{x}{x-2}, \quad\quad(3,3)$$

Step-by-Step Solution

Answer

The slope at \((3, 3)\) is \(-2\) and the tangent line equation is \(y = -2x + 9\).

Step 1: Differentiate the Function

To find the slope of the tangent line to the function's graph at the point \((3, 3)\), we need to find the derivative \(g'(x)\). The function is given as \(g(x) = \frac{x}{x-2}\). We will use the quotient rule, which states \(\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^2}\), where \(u = x\) and \(v = x-2\). First, find the derivatives of \(u\) and \(v\): \(u' = 1\) and \(v' = 1\). Substituting these into the quotient rule gives:\[g'(x) = \frac{(x-2)(1) - x(1)}{(x-2)^2} = \frac{x-2-x}{(x-2)^2} = \frac{-2}{(x-2)^2}\].

Step 2: Evaluate the Derivative at the Given Point

Substitute \(x = 3\) into the derivative \(g'(x)\) to find the slope of the tangent line at the point \((3, 3)\):\[g'(3) = \frac{-2}{(3-2)^2} = \frac{-2}{1} = -2\].So, the slope of the tangent line is \(-2\).

Step 3: Find the Tangent Line Equation

Now that we have the slope of the tangent line, \(m = -2\), and the point \((3, 3)\), we can use the point-slope form of a linear equation to write the equation of the tangent line. The point-slope form is \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is a point on the line. Substituting \((x_1, y_1) = (3, 3)\) and \(m = -2\):\[y - 3 = -2(x - 3)\].Simplify to get the equation of the tangent line:\[y = -2x + 9\].

Key Concepts

Derivative

The derivative is a fundamental concept in calculus, representing the rate of change of a function with respect to a variable. In simpler terms, it tells us how fast or slow a function is changing at any point along its curve. If you imagine driving a car and watching the speedometer, the derivative represents the speed at which you're traveling (the rate of change of distance with respect to time).To find a derivative:

• Identify the function you want to differentiate.
• Apply differentiation rules, such as the power rule, product rule, quotient rule, or chain rule, depending on the function's form.

In the given problem, we have the function \(g(x) = \frac{x}{x-2}\). To find its derivative, we'll employ the quotient rule, a handy tool when dealing with fractions of two functions. Remember, the outcome of this process — the derivative — will tell us the slope of the tangent line at any given point on the function's graph.

Tangent Line

A tangent line is a straight line that touches a function's curve at just one point. This line mirrors how the function behaves at that particular spot. Imagine gently resting a ruler on the curve so that it touches just one point, without cutting through the curve.The importance of a tangent line:

• It provides an approximation of the function near the point of tangency.
• Its slope at that point is equal to the derivative of the function at the same point.

In our task, after finding the derivative \( g'(x) \) and evaluating it at \( x = 3 \), we determined the slope of our tangent line as -2. Hence, using the point-slope form \( y - y_1 = m(x - x_1) \), where \((x_1, y_1)\) represents the point of tangency, we write the equation: \( y - 3 = -2(x - 3) \). Simplifying gives the tangent line equation, \( y = -2x + 9 \), illustrating the function's behavior around point \((3, 3)\).

Quotient Rule

The quotient rule, a crucial tool in calculus, helps differentiate functions where one function is divided by another. It's especially useful when you're dealing with a complex fraction. The rule is expressed as:\[\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^2}\]Here's how you apply it:

• Identify the numerator \(u\) and the denominator \(v\) of the function.
• Differentiate both \(u\) and \(v\).
• Substitute \(u\), \(v\), and their derivatives \(u'\), \(v'\) into the quotient rule formula.

In our exercise, \(u = x\) and \(v = x-2\), meaning \(u' = 1\) and \(v' = 1\). Plugging these into the quotient rule formula results in:\[g'(x) = \frac{(x-2)(1) - x(1)}{(x-2)^2} = \frac{-2}{(x-2)^2}\]This derivative helps us find the slope of the tangent line to the graph of the function at any point \(x\). In particular, it showed that at \(x=3\), the slope is \(-2\), leading us to the tangent line equation.