Problem 3
Question
Determine whether the function is continuous on the entire real line. Explain your reasoning. \(f(x)=\frac{1}{x^{2}-4}\)
Step-by-Step Solution
Verified Answer
The function \(f(x)=\frac{1}{x^{2}-4}\) is continuous on the real line except at \(x=-2\) and \(x=2\). The intervals of continuity are \((-∞, -2)\), \((-2, 2)\), and \((2, ∞)\).
1Step 1: Identify the discontinuity points
First, locate where the function \(f(x)=\frac{1}{x^{2}-4}\) becomes undefined. It becomes undefined when the denominator equals zero. So, solve the equation \(x^{2}-4=0\) for \(x\).
2Step 2: Solve the equation
Solving \(x^{2}-4=0\) gives \(x^{2}=4\), which implies \(x=\pm2\). The function \(f(x)\) is therefore undefined at \(x=-2\) and \(x=2\).
3Step 3: Establish the intervals of continuity
Since the function is undefined only at \(x=-2\) and \(x=2\), it is continuous everywhere else on the real line. Therefore, the intervals of continuity of \(f(x)\) are \((-∞, -2)\), \((-2, 2)\) and \((2, ∞)\).
Key Concepts
Discontinuity PointsUndefined FunctionIntervals of Continuity
Discontinuity Points
Discontinuity points in a function occur where the function is not continuous. These are specific values of the input where the function suddenly jumps, breaks, or stops being defined. In the provided example, the goal is to determine these points for the function \( f(x)=\frac{1}{x^{2}-4} \).
To find such points, you must identify where the denominator of the fraction equals zero, as a division by zero is undefined. Thus, solving the equation \( x^{2}-4=0 \) allows us to pinpoint these critical values. The solution to \( x^{2}=4 \) reveals two points: \( x=2 \) and \( x=-2 \).
At these values, the denominator becomes zero, rendering the function discontinuous at \( x=2 \) and \( x=-2 \). As a result, these are the discontinuity points for the given function.
To find such points, you must identify where the denominator of the fraction equals zero, as a division by zero is undefined. Thus, solving the equation \( x^{2}-4=0 \) allows us to pinpoint these critical values. The solution to \( x^{2}=4 \) reveals two points: \( x=2 \) and \( x=-2 \).
At these values, the denominator becomes zero, rendering the function discontinuous at \( x=2 \) and \( x=-2 \). As a result, these are the discontinuity points for the given function.
Undefined Function
When talking about an undefined function in a calculus context, it refers to places where a function fails to produce a value. This generally happens when an operation within the function can't be mathematically satisfied.
In our example, the function \( f(x)=\frac{1}{x^{2}-4} \) becomes undefined when \( x^{2}-4 \) equals zero due to division by zero. This occurs specifically at \( x=2 \) and \( x=-2 \).
Whenever the denominator in a fraction equals zero, the function is undefined at those x-values because division by zero is a mathematical operation that has no solution. Hence, the function does not exist at these points, which are crucial for understanding the behavior and limitations of the function.
In our example, the function \( f(x)=\frac{1}{x^{2}-4} \) becomes undefined when \( x^{2}-4 \) equals zero due to division by zero. This occurs specifically at \( x=2 \) and \( x=-2 \).
Whenever the denominator in a fraction equals zero, the function is undefined at those x-values because division by zero is a mathematical operation that has no solution. Hence, the function does not exist at these points, which are crucial for understanding the behavior and limitations of the function.
Intervals of Continuity
Intervals of continuity describe the range of x-values where a function is continuous, avoiding any undefined points or breaks. For continuous intervals, you don't encounter any holes, jumps, or asymptotic behavior.
In the case of the function \( f(x)=\frac{1}{x^{2}-4} \), continuity is interrupted only at \( x=-2 \) and \( x=2 \), where the function is undefined. Consequently, the function remains continuous on the following intervals:
In the case of the function \( f(x)=\frac{1}{x^{2}-4} \), continuity is interrupted only at \( x=-2 \) and \( x=2 \), where the function is undefined. Consequently, the function remains continuous on the following intervals:
- \((-∞, -2)\)
- \((-2, 2)\)
- \((2, ∞)\)
Other exercises in this chapter
Problem 3
Use a graphing utility to graph the function and find its average rate of change on the interval. Compare this rate with the instantaneous rates of change at th
View solution Problem 3
Find the slope of the tangent line to \(y=x^{n}\) at the point \((1,1)\). (a) \(y=x^{-1}\) (b) \(y=x^{-1 / 3}\)
View solution Problem 3
Complete the table and use the result to estimate the limit. Use a graphing utility to graph the function to confirm your result. $$ \lim _{x \rightarrow 2} \fr
View solution Problem 4
Identify the inside function, \(u=g(x)\), and the outside function, \(y=f(u)\). $$ y=f(g(x)) \quad u=g(x) \quad y=f(u) $$ $$ y=\left(x^{2}+1\right)^{4 / 3} $$
View solution