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Let \(f(x)=\frac{1}{x-1}\). a. Find the derivative \(f^{\prime}\) of \(f\). b. Find an equation of the tangent line to the curve at the point \(\left(-1,-\frac{1}{2}\right)\). c. Sketch the graph of \(f\) and the tangent line to the curve at the point \(\left(-1,-\frac{1}{2}\right)\).

Short Answer

Expert verified
a. The derivative of \(f(x)=\frac{1}{x-1}\) is found using the quotient rule, resulting in \(f^{\prime}(x) = -\frac{1}{(x-1)^2}\). b. The equation of the tangent line at the point \((-1, -\frac{1}{2})\) is given by \(y = -\frac{1}{4}x - \frac{3}{4}\). c. To sketch the graph of \(f(x)\) and the tangent line, draw the vertical asymptote at \(x=1\), plot the function \(f(x)=\frac{1}{x-1}\), plot the point \((-1, -\frac{1}{2})\), and draw the tangent line with the equation \(y = -\frac{1}{4}x - \frac{3}{4}\).

Step by step solution

01

Determine the derivative of \(f(x)\) by using the quotient rule.

Use the quotient rule to find the derivative of \(f(x) = \frac{1}{x-1}\). Recall that the quotient rule states that if \(y = \frac{u}{v}\), then \(y' = \frac{u'v - uv'}{v^2}\), where \(u'\) and \(v'\) are the derivatives of \(u\) and \(v\), respectively. In this case, \(u = 1\) and \(v = x - 1\), so \(u' = 0\) and \(v' = 1\). Using the quotient rule, we find: \[f^{\prime}(x) = \frac{(0)(x-1) - (1)(1)}{(x-1)^2} = -\frac{1}{(x-1)^2}\] #Second Step: Equation of the Tangent Line#
02

Determine the equation of the tangent line at the point \(\left(-1,-\frac{1}{2}\right)\).

To find the equation of the tangent line at the point \(\left(-1,-\frac{1}{2}\right)\), we need to first evaluate the derivative at \(-1\) to find the slope of the tangent line: \[f^{\prime}(-1) = -\frac{1}{(-1-1)^2} = -\frac{1}{4}\] Now, use the point-slope form of a linear equation, given by \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is a point on the line and \(m\) is the slope. In this case, \((x_1, y_1) = \left(-1, -\frac{1}{2}\right)\) and \(m = -\frac{1}{4}\): \[y + \frac{1}{2} = -\frac{1}{4}(x+1)\] Finally, to put this into slope-intercept form (\(y = mx + b\)), we get: \[y = -\frac{1}{4}x - \frac{3}{4}\] #Third Step: Sketch the Graph of \(f(x)\) and the Tangent Line#
03

Draw the graph of \(f(x) = \frac{1}{x-1}\) and the tangent line \(y = -\frac{1}{4}x - \frac{3}{4}\).

To sketch the graph of \(f(x) = \frac{1}{x-1}\) and the tangent line \(y = -\frac{1}{4}x - \frac{3}{4}\), follow these steps: 1. Draw the vertical asymptote at \(x = 1\) since the function is undefined there. 2. Draw the graph of the function \(f(x) = \frac{1}{x-1}\). Note that as \(x\) approaches the asymptote, the function goes towards infinity or negative infinity. 3. Plot the point \(\left(-1, -\frac{1}{2}\right)\) on the graph. 4. Draw the tangent line passing through the point \(\left(-1, -\frac{1}{2}\right)\) with the slope \(-\frac{1}{4}\) and the equation \(y = -\frac{1}{4}x - \frac{3}{4}\). By doing this, you should have a graph with the given function and tangent line, which are crossing at the given point.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative of a Function
In calculus, the derivative of a function is a fundamental concept that measures how a function’s output changes as its input changes. It is often represented as \( f'(x) \) or \( \frac{df}{dx} \) . The derivative helps us understand the rate of change at a particular point on a graph, which is the slope of the tangent line to the function at that point.

For example, to find the derivative of a simple power function like \( f(x) = x^n \), we use the power rule which states that \( f'(x) = nx^{n-1} \). However, if a function is a ratio of two different functions (a quotient), such as \( f(x) = \frac{1}{x-1} \), we must use the quotient rule to find its derivative, which leads us to our next critical topic.
Quotient Rule Differentiation
The quotient rule is a technique in calculus used to find the derivative of functions that are ratios of two differentiable functions. The formula for the quotient rule is given by

\[ f'(x) = \frac{u'v - uv'}{v^2} \]

where \( u \) and \( v \) are functions of \( x \), and \( u' \) and \( v' \) are their respective derivatives. When applying the quotient rule to the function \( f(x) = \frac{1}{x-1} \), we identify \( u = 1 \) and \( v = x - 1 \), and their derivatives \( u' = 0 \) since 1 is a constant, and \( v' = 1 \) since the derivative of \( x \) is 1.

Applying the quotient rule as shown in the textbook solution allows us to find the derivative that is needed to determine the slope of a tangent line at a specific point on the function's graph.
Point-Slope Form of a Linear Equation
The point-slope form of a linear equation is a way to write the equation of a line when you know the slope and a point through which the line passes. It is written as:

\[ y - y_1 = m(x - x_1) \]

where \((x_1, y_1)\) is the point on the line, and \(m\) is the slope of the line. In our case, we have the point \((-1, -1/2)\) and the slope \( -1/4 \), calculated as the value of the derivative at \(x_1 = -1\).

This form is particularly useful because it’s straightforward to plug in the values and quickly derive the equation of the tangent line, as was demonstrated in the solution steps for finding the equation of the tangent line to the curve at the given point.
Sketching Graphs of Functions
When sketching the graph of a function, we use the function’s properties, such as intercepts, asymptotes, and intervals of increase and decrease, to create an accurate representation of the function's behavior. Recognizing these features and understanding the behavior of a function near asymptotes are pivotal in sketching.

For the function \( f(x) = \frac{1}{x-1} \), we know there is a vertical asymptote at \(x = 1\) since the function is undefined when \( x - 1 = 0 \). The graph tends towards infinity as it approaches the asymptote from the left side and negative infinity from the right side.

Additionally, the tangent line, which we found by using the function’s derivative and the point-slope form, is drawn at the specific point we are interested in. This line just touches the function's graph at one point and its slope tells us the rate of change of the function at that specific point. This visual combination on the same set of axes provides a rich understanding of how the function behaves and changes — step 3 in our solution showcases this process.

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