Problem 17
Question
In Problems 15-24, find the values of \(x \in\) R for which the given functions are both defined and continuous. $$ f(x)=\frac{x^{2}+1}{x-1} $$
Step-by-Step Solution
Verified Answer
The function is defined and continuous for \( x \in \mathbb{R} \setminus \{1\} \).
1Step 1: Identify undefined points
To determine where the function \( f(x) = \frac{x^2 + 1}{x - 1} \) is undefined, set the denominator equal to zero and solve for \( x \). The denominator is zero when \( x - 1 = 0 \), which gives \( x = 1 \). Thus, the function is undefined at \( x = 1 \).
2Step 2: Determine continuity
A rational function like \( f(x) = \frac{x^2 + 1}{x - 1} \) is continuous at all points except where it is undefined. Since we found in Step 1 that the function is undefined at \( x = 1 \), the function is discontinuous at this point. Otherwise, it is continuous for all other real values of \( x \).
3Step 3: Solution summary
The function \( f(x) = \frac{x^2 + 1}{x - 1} \) is both defined and continuous for all real numbers \( x \) except \( x = 1 \). This means \( x \in \mathbb{R} \setminus \{1\} \).
Key Concepts
Undefined Points in FunctionsDiscontinuityRational Functions
Undefined Points in Functions
When working with mathematical functions, it's important to figure out where a function may be undefined. A function can become undefined when you encounter certain conditions that make its computation impossible or undefined. For instance, in the function \( f(x) = \frac{x^2 + 1}{x - 1} \), setting up the denominator \( x - 1 \) equal to zero can lead us to these points. Solving for \( x \), you find that at \( x = 1 \) the denominator equals zero, making the function undefined at this point.
- Undefined points arise where the denominator of a rational function turns zero.
- Once you find these points, the function cannot evaluate at these values of \( x \).
Discontinuity
Discontinuity refers to a point or collection of points in the domain of a function where the function is not continuous. For rational functions, these discontinuities typically occur at the undefined points. In our function \( f(x) = \frac{x^2 + 1}{x - 1} \), the only point where it is discontinuous is \( x = 1 \).
- At \( x = 1 \), the function jumps from being defined to undefined, thus creating a break in the continuity.
- Beyond this point, the function continues without interruption over other real values of \( x \).
Rational Functions
Rational functions are mathematical expressions represented as a ratio of two polynomials. In the expression \( f(x) = \frac{x^2 + 1}{x - 1} \), the numerator is \( x^2 + 1 \) and the denominator is \( x - 1 \). These functions have some intriguing characteristics due to their form:
- They are continuous across most of their domain, starkly excepted where the denominator zeroes out.
- The denominator can create vertical asymptotes, creating breaks in the graph where the function shoots off towards infinity.
- The degree of polynomials affects the behavior – for instance, more complex polynomials might introduce additional asymptotes or undefined points.
Other exercises in this chapter
Problem 17
17\. LL(a) Use a graphing calculator to sketch the graph of \(y=f(x)\). (b) Show that \(-x^{2} \leq x^{2} \cos \frac{1}{x} \leq x^{2}\) holds for \(x \neq 0\).
View solution Problem 17
For each of the following equations show that the equation has a \mathrm{\\{} \text { root in the given interval. Then use the bisection search method, } implem
View solution Problem 17
Evaluate the limits. $$ \lim _{x \rightarrow-\infty} \exp [x] $$
View solution Problem 17
In Problems 1-32, use a table or a graph to investigate each limit. $$ \lim _{x \rightarrow 0} \frac{1}{1+x^{2}} $$
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