Problem 32
Question
In Problems , graph each function and, on the basis of the graph, guess where the function is not differentiable. (Assume the largest possible domain.)$$ y=\frac{x-1}{x+1} $$
Step-by-Step Solution
Verified Answer
The function \( y = \frac{x-1}{x+1} \) is not differentiable at \( x = -1 \).
1Step 1: Simplify the Function Expression
The given function is \( y = \frac{x-1}{x+1} \). This is a rational function where the numerator is \( x-1 \) and the denominator is \( x+1 \). We need to find where this function can be undefined or not differentiable, which typically occurs at points where the denominator is zero.
2Step 2: Identify Points of Discontinuity
A function is not differentiable at points of discontinuity. Set the denominator of the function \( x+1 = 0 \), to find the points of discontinuity. Solving this gives \( x = -1 \). So the function is not defined and hence not differentiable at \( x = -1 \).
3Step 3: Graph the Function
To understand the nature of the function, sketch a graph of \( y = \frac{x-1}{x+1} \). This will visually confirm the behavior near \( x = -1 \). The graph will have a vertical asymptote at \( x = -1 \), confirming the non-differentiability at this point.
4Step 4: Analyze the Function Further
Verify that \( x = -1 \) is the only point where the function is discontinuous. Other than this, for all other values of \( x \), the function is continuous and differentiable as it is a ratio of two polynomials without issues other than at \( x = -1 \).
Key Concepts
Rational FunctionsPoints of DiscontinuityVertical Asymptote
Rational Functions
Rational functions are fascinating elements of calculus. They take the form \( f(x) = \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials. The degrees of these polynomials can affect the behavior and properties of the function. The behavior of these functions is greatly influenced by the denominator—since if the denominator becomes zero at any point, it can make the function undefined.
These types of functions can produce complex graphs displaying characteristics like horizontal asymptotes, vertical asymptotes, or even holes. Understanding rational functions provides insight into their limits, asymptotes, and the continuity of graphs.
Since rational functions include divisibility, identifying where they are undefined or not differentiable is crucial. The rational function given, \( y = \frac{x-1}{x+1} \), requires focusing on the denominator to find possible discontinuities. Here, the fraction becomes undefined at \( x = -1 \), establishing a pivotal point of interest.
These types of functions can produce complex graphs displaying characteristics like horizontal asymptotes, vertical asymptotes, or even holes. Understanding rational functions provides insight into their limits, asymptotes, and the continuity of graphs.
Since rational functions include divisibility, identifying where they are undefined or not differentiable is crucial. The rational function given, \( y = \frac{x-1}{x+1} \), requires focusing on the denominator to find possible discontinuities. Here, the fraction becomes undefined at \( x = -1 \), establishing a pivotal point of interest.
Points of Discontinuity
In calculus, points of discontinuity are locations on the function's graph where the function is not continuous. They are crucial because they represent potential places where the function is not differentiable.
To find these points in a rational function, examine the denominator to identify where it equals zero, which causes the function to become undefined. For instance, in the function \( y = \frac{x-1}{x+1} \), setting the denominator \( x+1 = 0 \), solves to \( x = -1 \). Hence, \( x = -1 \) is a point of discontinuity.
Points of discontinuity in rational functions often appear as
To find these points in a rational function, examine the denominator to identify where it equals zero, which causes the function to become undefined. For instance, in the function \( y = \frac{x-1}{x+1} \), setting the denominator \( x+1 = 0 \), solves to \( x = -1 \). Hence, \( x = -1 \) is a point of discontinuity.
Points of discontinuity in rational functions often appear as
- Vertical asymptotes
- Removable discontinuities (holes)
Vertical Asymptote
A vertical asymptote represents a line that a graph approaches but never crosses or touches. In a rational function, vertical asymptotes occur at values for which the denominator is zero, but the numerator is not simultaneously zero. These special lines highlight the infinite nature of the function's growth or decay near particular points.
For the rational function \( y = \frac{x-1}{x+1} \), the vertical asymptote appears at \( x = -1 \). As \( x \) approaches \( -1 \), the function approaches infinity or negative infinity, but it never quite meets the line \( x = -1 \). This results in a division in the graph and demonstrates why \( x = -1 \) is a point where the function is not differentiable.
Understanding vertical asymptotes is crucial for analyzing the complete behavior of a graph. Key points include:
For the rational function \( y = \frac{x-1}{x+1} \), the vertical asymptote appears at \( x = -1 \). As \( x \) approaches \( -1 \), the function approaches infinity or negative infinity, but it never quite meets the line \( x = -1 \). This results in a division in the graph and demonstrates why \( x = -1 \) is a point where the function is not differentiable.
Understanding vertical asymptotes is crucial for analyzing the complete behavior of a graph. Key points include:
- Marking undefined values in a function.
- Indicating places where the graph shoots indefinitely up or down.
Other exercises in this chapter
Problem 32
Differentiate $$ f(x)=\frac{3(x-1)^{2}}{2+a} $$ with respect to \(x\). Assume that \(a\) is a positive constant.
View solution Problem 32
Differentiate $$ f(N)=\frac{b N^{2}+N}{K+b} $$ with respect to \(N\). Assume that \(b\) and \(K\) are positive constants.
View solution Problem 32
Differentiate $$ g(N)=\frac{N}{(k+b N)^{2}} $$ with respect to \(N\). Assume that \(b\) and \(k\) are positive constants.
View solution Problem 33
Suppose that the specific growth rate of a plant is \(1 \%\); that is, if \(B(t)\) denotes the biomass at time \(t\), then $$\frac{1}{B(t)} \frac{d B}{d t}=0.01
View solution