Problem 8
Question
Find the slope of the tangent line to the graph of \(f\) at the given point. $$f(x)=\frac{6}{x+1} \quad \text { at }(2,2)$$
Step-by-Step Solution
Verified Answer
The slope of the tangent line at (2,2) is \(-\frac{2}{3}\).
1Step 1: Understand the Problem
We are asked to find the slope of the tangent line to the graph of the function \(f(x) = \frac{6}{x+1}\) at a specific point, \((2, 2)\). The slope of the tangent line can be found by evaluating the derivative of \(f(x)\) at the given point \(x = 2\).
2Step 2: Differentiate the Function
To find the slope of the tangent line, we first need to find the derivative of \(f(x)\). Using the power rule or simply by recognizing this as a simple quotient with a constant numerator, the derivative \(f'(x)\) of \(f(x) = \frac{6}{x+1}\) can be calculated as follows: \[f'(x) = -\frac{6}{(x+1)^2}\].
3Step 3: Evaluate the Derivative at the Given Point
Now that we have the expression for the derivative, \(f'(x) = -\frac{6}{(x+1)^2}\), we substitute \(x = 2\) into this derivative to find the slope at the given point:\[f'(2) = -\frac{6}{(2+1)^2} = -\frac{6}{9} = -\frac{2}{3}\].
4Step 4: Conclusion
The slope of the tangent line to the curve \(f(x) = \frac{6}{x+1}\) at the point \((2, 2)\) is \(-\frac{2}{3}\).
Key Concepts
DifferentiationCalculusDerivative Evaluation
Differentiation
Differentiation is a fundamental concept in calculus that deals with finding the rate at which a function is changing at any given point. This rate of change is known as the derivative. To differentiate a function means to compute its derivative, which gives you the slope of the tangent line to the graph of the function at any point. In simpler terms, it tells you how steep the graph is at that particular spot.
When you see a function like \( f(x) = \frac{6}{x+1} \), differentiation involves getting from this expression to another one, \( f'(x) = -\frac{6}{(x+1)^2} \), which describes that slope for every \( x \).
To find this, you break down the function into parts you can handle, often using rules like the power rule or recognizing simple patterns, like a quotient here. Once you have the derivative, you're ready to move on to more specific calculations involving it.
When you see a function like \( f(x) = \frac{6}{x+1} \), differentiation involves getting from this expression to another one, \( f'(x) = -\frac{6}{(x+1)^2} \), which describes that slope for every \( x \).
To find this, you break down the function into parts you can handle, often using rules like the power rule or recognizing simple patterns, like a quotient here. Once you have the derivative, you're ready to move on to more specific calculations involving it.
Calculus
Calculus is a major branch of mathematics focused on change and motion. It covers two main concepts: differentiation and integration. While differentiation looks at how things change (like slopes of tangent lines, as in our example), integration accumulates values to find things like areas under curves.
In our exercise, we're zooming in on the differentiation part of calculus. Here, this has involved taking a specific function, \( f(x) = \frac{6}{x+1} \), and figuring out how it behaves at precise points, specifically at \( x = 2 \). This is a small but important piece of how calculus allows us to understand and model real-world phenomena, whether they involve motion, area, or growth.
In our exercise, we're zooming in on the differentiation part of calculus. Here, this has involved taking a specific function, \( f(x) = \frac{6}{x+1} \), and figuring out how it behaves at precise points, specifically at \( x = 2 \). This is a small but important piece of how calculus allows us to understand and model real-world phenomena, whether they involve motion, area, or growth.
Derivative Evaluation
Evaluating a derivative involves substituting a specific value into the derivative of a function to find the slope of the tangent line at that point. This step is crucial if you want to understand the exact behavior of a function at a specific location on its graph.
For the function \( f(x) = \frac{6}{x+1} \), after finding the derivative \( f'(x) = -\frac{6}{(x+1)^2} \), the next task is to plug in \( x = 2 \). This gives us \( f'(2) = -\frac{6}{9} = -\frac{2}{3} \).
What this process shows is how well these steps of finding and evaluating derivatives make it possible to precisely determine the slope of the tangent line. This slope indicates the immediate rate of change of the function at \( x = 2 \), which concludes this particular exercise.
For the function \( f(x) = \frac{6}{x+1} \), after finding the derivative \( f'(x) = -\frac{6}{(x+1)^2} \), the next task is to plug in \( x = 2 \). This gives us \( f'(2) = -\frac{6}{9} = -\frac{2}{3} \).
What this process shows is how well these steps of finding and evaluating derivatives make it possible to precisely determine the slope of the tangent line. This slope indicates the immediate rate of change of the function at \( x = 2 \), which concludes this particular exercise.
Other exercises in this chapter
Problem 8
Evaluate the limit and justify each step by indicating the appropriate Limit Law(s). $$\lim _{x \rightarrow 1}\left(\frac{x^{4}+x^{2}-6}{x^{4}+2 x+3}\right)^{2}
View solution Problem 8
Find the limit. $$\lim _{x \rightarrow \infty} \frac{2-3 x}{4 x+5}$$
View solution Problem 9
(a) Estimate the area under the graph of \(f(x)=1 / x\) from \(x=1\) to \(x=5\) using four approximating rectangles and right endpoints. Sketch the graph and th
View solution Problem 9
Evaluate the limit and justify each step by indicating the appropriate Limit Law(s). $$\lim _{t \rightarrow-2}(t+1)^{9}\left(t^{2}-1\right)$$
View solution