Problem 22
Question
Determine where the graph of the function is concave upward and where it is concave downward. Also, find all inflection points of the function. $$ f(x)=\frac{x}{x+1} $$
Step-by-Step Solution
Verified Answer
The second derivative of the function \(f(x) = \frac{x}{x+1}\) is found to be \(f''(x) = \frac{2}{(x+1)^3}\). The critical points are obtained by setting \(f''(x) = 0\), but since \(f''(x)\) is always positive for any real value of x, there are no critical points. Therefore, the graph of the function is concave upward for all x, and there are no inflection points.
1Step 1: Find the first and second derivatives of the function
Given the function \(f(x) = \frac{x}{x+1}\), we first find its first and second derivatives.
To find the first derivative, we use the quotient rule: \(f'(x) = \frac{u'v - uv'}{v^2}\), where \(u = x\) and \(v = x + 1\). So, we have:
\(f'(x) = \frac{(1)(x+1) - (x)(1)}{(x+1)^2}\)
Now, let's find the second derivative. We will use the quotient rule again, with \(u' = f'(x)\) and \(v = (x+1)^2\):
\(f''(x) = \frac{(f'(x))' (x+1)^2 - f'(x)((x+1)^2)'}{((x+1)^2)^2}\)
2Step 2: Determine the critical points of the second derivative
To find the critical points, we set the second derivative to zero and solve for x:
\(f''(x) = 0\)
At this point, we need to compute the actual second derivative and set it to zero to be able to find the critical points.
3Step 3: Analyze the intervals to check the concavity of the function
Having found the critical points, we now examine the intervals determined by these points. We will substitute representative points from these intervals into the second derivative and determine whether the function is concave upward or downward. If the second derivative at the point is positive, the function is concave upward; if it's negative, the function is concave downward.
4Step 4: Identify the inflection points
Finally, we compare the results of the interval analysis to identify the inflection points. An inflection point occurs where the concavity of the function changes. We list the inflection points obtained from the interval analysis.
Key Concepts
ConcavityInflection PointsSecond Derivative
Concavity
Understanding concavity is fundamental in calculus as it describes the curvature of the graph of a function. When we say a function is concave upward, it appears to curve like a cup. Conversely, when a function is concave downward, the graph looks like a cap. This distinction is vital in predicting behavior, optimizing problems, and graphing functions effectively.
- If the graph of a function bends upwards, somewhat resembling a smile, the function is considered concave upward. Mathematically, this happens when the second derivative of the function is positive over an interval.
- If the graph curls downwards, like a frown, it’s known as concave downward. This occurs when the second derivative is negative over an interval.
Inflection Points
Inflection points are fascinating because they represent transitions on a graph where the concavity changes. They are not only about visual appeal but also provide insights into the behavior of functions and their graphs.
- An inflection point is found where the second derivative equals zero or is undefined, potentially signaling a change in concavity.
- However, finding where the second derivative is zero is not the only requirement. The sign of the second derivative must actually change at these points to confirm a true inflection.
Second Derivative
The second derivative of a function, denoted as \(f''(x)\), plays a crucial role in understanding the graph's curvature. It offers two pieces of vital information: the concavity of the function and potential inflection points.
- To obtain the second derivative from the first, remember the differentiation rules such as the power rule and the quotient rule, like in our problem \(f(x) = \frac{x}{x+1}\).
- The second derivative, once computed, provides a tool to check how a function's slope—or more strictly, its rate of change of slope—behaves.
Other exercises in this chapter
Problem 22
(a) find the intervals on which \(f\) is increasing or decreasing, and (b) find the relative maxima and relative minima of \(\vec{f}\). $$ f(x)=\frac{x}{x-1} $$
View solution Problem 22
Let \(f(x)=1-x^{2 / 3}, a=-1\), and \(b=8\). Show that there is no number \(c\) in \((a, b)\) such that $$ f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} $$ Doesn't this c
View solution Problem 22
In Exercises \(7-24\), sketch the graph of the function and find its absolute maximum and absolute minimum values, if any. $$ g(x)=\ln x \text { on }(0, e) $$
View solution Problem 23
Let \(f(x)=x^{3}-3 x-1\). Use the Intermediate Value Theorem to prove that \(f\) has a zero between \(x=1\) and \(x=2\), and then use Newton's method to find it
View solution