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Chapter 2: Problem 156

Let \(f(x)=\frac{x^{4}-1}{x^{2}-1}\) for \(x \neq-1,1\). a. Sketch the graph of \(f\). b. Is it possible to find values \(k_{1}\) and \(k_{2}\) such that \(f(-1)=k\) and \(f(1)=k_{2},\) and that makes \(f(x)\) continuous for all real numbers? Briefly explain.

Short Answer

Expert verified
Yes, set \( k_1 = 2 \) and \( k_2 = 2 \) to make \( f(x) \) continuous everywhere.

Step by step solution

01

Simplify the Function

The function is given as \( f(x)=\frac{x^{4}-1}{x^{2}-1} \). First, simplify the expression \( \frac{x^4 - 1}{x^2 - 1} \). Notice that \( x^4 - 1 \) can be factored as \((x^2-1)(x^2+1)\). This allows us to rewrite the function as:\[f(x) = \frac{(x^2 - 1)(x^2 + 1)}{x^2 - 1}\]For \( x eq -1, 1 \), this simplifies to:\[f(x) = x^2 + 1\] because the \(x^2 - 1\) terms cancel out.
02

Sketch the Graph

The simplified function \( f(x) = x^2 + 1 \) is a parabola opening upwards, shifted 1 unit upwards from the standard \( x^2 \). The graph has the shape of a "U" with a vertex at (0, 1). To accurately sketch it, note that asymptotically for large \(x\), the graph behaves like \(y = x^2\). It is continuous for \(x eq -1, 1\).
03

Analyzing Continuity at Problem Points

Since \( f(x) = x^2 + 1 \) for \( x eq -1, 1 \), we must consider continuity at \( x = -1 \) and \( x = 1 \). At these points, the original function form \( \frac{x^4 - 1}{x^2 - 1} \) is undefined, but we need to find potential values \( k_1 \) and \( k_2 \) to make it continuous.
04

Determine \( f(-1) \) and \( f(1) \) for Continuity

To ensure continuity, the limit of \( f(x) \) as \( x \) approaches -1 from both sides must equal \( f(-1) \) and similarly for \( x = 1 \). Calculate:\[ \lim_{x \to -1} f(x) = (-1)^2 + 1 = 2 \]Thus, set \( f(-1) = k_1 = 2 \).Similarly, for \( x = 1 \):\[ \lim_{x \to 1} f(x) = 1^2 + 1 = 2 \]Thus, set \( f(1) = k_2 = 2 \).
05

Conclusion

By defining \( f(-1) = 2 \) and \( f(1) = 2 \), the function \( f(x) \) is continuous over all real numbers. At these specific points \( x = -1 \) and \( x = 1 \), the original issue of undefined values is resolved by specifying these outputs based on the limits of the simplified function.

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Key Concepts

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

Function Continuity
In calculus, the concept of continuity for a function is crucial. A function is considered continuous at a particular point if there are no abrupt changes or gaps at that point. More formally, a function \( f(x) \) is continuous at a point \( c \) if the following conditions are satisfied:
  • The function \( f \) is defined at \( c \); meaning \( f(c) \) exists.
  • The limit of the function as it approaches \( c \) from both sides exists: \( \lim_{x \to c} f(x) \).
  • The limit of the function as it approaches \( c \) from both sides is equal to the value of the function at that point: \( \lim_{x \to c} f(x) = f(c) \).
In our exercise involving \( f(x) = \frac{x^4 - 1}{x^2 - 1} \), continuity at \( x = -1 \) and \( x = 1 \) is achieved by redefining \( f \) at these points to equal the limits of the simplified function as described in the steps. This ensures there are no jumps or breaks at these points, making \( f(x) \) continuous across all real numbers.
Graph Sketching
Graph sketching is an essential skill in calculus, which involves plotting the behavior of a function across its domain. For the function \( f(x) = x^2 + 1 \), after simplification, the graph is a smooth parabola that opens upwards. Sketching it involves the following steps:
  • Identify the vertex of the parabola, which for \( f(x) = x^2 + 1 \) is at the point (0, 1).
  • Determine the direction the parabola opens. In this case, since the coefficient before \( x^2 \) is positive, it opens upwards.
  • Understand the general shape and orientation: for large values of \( x \), the behavior mirrors the traditional \( y = x^2 \) parabola, indicating symmetry about the y-axis.
Using this layout, the graph helps visualize where the function is not defined, and how the redefined points at \( x = -1 \) and \( x = 1 \) fit seamlessly into the overall shape, maintaining continuity.
Limit Evaluation
Evaluating limits involves finding the value that a function approaches as the input approaches a particular point. This is particularly significant when a function is not originally defined at that point. In the original function \( f(x) = \frac{x^4 - 1}{x^2 - 1} \), it is undefined at \( x = -1 \) and \( x = 1 \). However, through simplification to \( f(x) = x^2 + 1 \), evaluating the limits is straightforward:
  • As \( x \) approaches \( -1 \), calculate \( \lim_{x \to -1} f(x) = (-1)^2 + 1 = 2 \).
  • As \( x \) approaches \( 1 \), calculate \( \lim_{x \to 1} f(x) = 1^2 + 1 = 2 \).
By setting \( f(-1) = 2 \) and \( f(1) = 2 \), these limits ensure the function is continuous at these points. Evaluating limits in this manner not only aids in resolving issues of undefined points but also plays a critical role in understanding the overall behavior and continuity of a function.

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Most popular questions from this chapter

In the following exercises, use the precise definition of limit to prove the given limits. $$\lim _{x \rightarrow 3} \frac{x^{2}-9}{x-3}=6$$

Using precise definitions of limits, prove that $$\lim _{x \rightarrow 0} f(x)$$ does not exist: \(f(x)=\left\\{\begin{array}{l}{1 \text { if } x \text { is rational }} \\ {0 \text { if } x \text { is irrational }}\end{array}\right.\) (Hint: Think about how you can always choose a rational number \(0< r< d,\) but \(|f(r)-0|=1.\))

Using precise definitions of limits, prove that $$\lim _{x \rightarrow 0} f(x)$$ does not exist, given that \(f(x)\) is the ceiling function. (Hint: Try any \(\delta< 1 . )\)

The following graph of the function \(f\) satisfies $$\lim _{x \rightarrow 3} f(x)=-1 .$$ In the following exercises, determine a value of \(\delta>0\) that satisfies each statement. If \(0<|x-3|<\delta,\) then \(|f(x)+1|<1\)

In the following exercises, sketch the graph of a function with the given properties. $$ \begin{array}{l}{\lim _{x \rightarrow-\infty} f(x)=0, \lim _{x \rightarrow-1^{-}} f(x)=\infty, \lim _{x \rightarrow-1^{+}} f(x)=-\infty} \\\ {f(0)=-1, \lim _{x \rightarrow 1^{-}} f(x)=-\infty, \lim _{x \rightarrow 1^{+}} f(x)=\infty, \lim _{x \rightarrow \infty} f(x)=0}\end{array} $$
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