Problem 104 - Calculus Early Transcendentals

Question

Let \(f(x)=\frac{|x|}{x} .\) Then \(f(-2)=-1\) and \(f(2)=1 .\) Therefore \(f(-2)<0

Step-by-Step Solution

Answer

In summary, the function \(f(x) = \frac{|x|}{x}\) is not continuous on the interval \([-2, 2]\) due to being undefined at \(x=0\). Since the Intermediate Value Theorem requires the function to be continuous on the closed interval, the fact that there is no value of \(c\) between -2 and 2 for which \(f(c)=0\) does not violate the Intermediate Value Theorem. Therefore, the IVT does not apply to this function on the given interval.

Step 1: Understand the function

The given function is \(f(x)=\frac{|x|}{x}\). We can see that \(f(x)\) takes the form \(-1\) for negative x-values, and \(1\) for positive x-values, as it divides the absolute value of \(x\) by \(x\) itself.

Step 2: Given values of the function

We are given that \(f(-2)=-1\) and \(f(2)=1\). As expected, the function returns these values as the sign of the input is preserved due to the division by \(x\) itself.

Step 3: Recall the conditions of the Intermediate Value Theorem

The Intermediate Value Theorem (IVT) states that if a function \(f\) is continuous on the closed interval \([a, b]\), and \(k\) is any number between \(f(a)\) and \(f(b)\), then there is at least one value \(c\) in the open interval \((a, b)\) such that \(f(c)=k\).

Step 4: Analyze the continuity

In order to determine if the IVT applies, we must analyze the continuity of the function \(f(x)=\frac{|x|}{x}\). This function is defined for all \(x \neq 0\), and \(f(x)\) is equal to \(-1\) for negative values of \(x\) and \(1\) for positive values of \(x\). However, at \(x=0\), the function is undefined, as division by zero is not allowed. So, the function is not continuous on the interval \([-2, 2]\), since it is undefined at \(x=0\).

Step 5: Determine if the IVT is violated

Since the Intermediate Value Theorem requires that the function is continuous on the closed interval, and our function \(f(x)=\frac{|x|}{x}\) is not continuous on \([-2, 2]\), the fact that there is no value of \(c\) between -2 and 2 for which \(f(c)=0\) does not violate the Intermediate Value Theorem. The IVT does not apply to this function on the given interval because of its lack of continuity.

Key Concepts

Continuity

In the realm of calculus, continuity plays a crucial role in understanding the behavior of functions. A function is said to be continuous at a point if there are no sudden jumps or breaks. This means that the graph of the function can be drawn without lifting your pencil from the paper. The function should smoothly transition from one point to the next.
For a function to be continuous over an interval, it should be continuous at every point within that interval. However, if there is even one point where the function is undefined or jumps abruptly, the function is not continuous over that interval.
For instance, with the function considered in the exercise, \( f(x) = \frac{|x|}{x} \), it is undefined at \( x = 0 \). This point of undefined behavior results in a discontinuity on the interval \([-2, 2]\).

Piecewise Function

Piecewise functions are those that have different expressions or "pieces" depending on the value of the input. They are a combination of several sub-functions, each defined over certain parts of the function's domain.
In the problem at hand, \( f(x) = \frac{|x|}{x} \) acts like a piecewise function in disguise. For negative values of \( x \), the function simplifies to \( -1 \), while for positive values, it simplifies to \( 1 \).
This behavior is akin to specifying \(-1\) for \( x < 0 \) and \( 1 \) for \( x > 0 \). The pieces connect the dots of different segments of the graph, each corresponding to their respective inputs. Understanding how piecewise functions operate is vital for analyzing their continuity. In this case, the abrupt change at \( x=0 \) prevents the function from being fully continuous.

Absolute Value

The concept of absolute value is essential when analyzing functions like \( f(x) = \frac{|x|}{x} \). The absolute value of a number is its distance from zero on the number line, regardless of direction. This means \( |x| \) is always non-negative.
Mathematically, \(|x| = x\) if \(x\) is non-negative, and \(|x| = -x\) if \(x\) is negative. This change in behavior depending on the sign of \( x \) is key to understanding how \(\frac{|x|}{x} \) morphs from \(-1\) to \(1\) as \(x\) crosses zero.
The absolute value function induces a transformation that removes the sign of \( x \), simplifying the analysis of functions that rely on positive versus negative outcomes. Hence, it influences the analysis of continuity and piecewise behavior.

Discontinuity

Discontinuities occur when there are breaks or jumps in the graph of a function. With \( f(x) = \frac{|x|}{x} \), a clear discontinuity is present at \( x = 0 \). The function dramatically changes its value from \(-1\) to \(1\) as \( x \) passes through zero, creating a jump discontinuity.
At the point of discontinuity, the function cannot be defined smoothly from one side to the other. This makes \( x = 0 \) a crucial breakpoint where the Intermediate Value Theorem does not hold, since the function isn’t continuous over the interval.
In calculus, recognizing different types of discontinuities helps in understanding where and why a function fails the conditions needed for certain theorems. In this case, the absence of continuity at \( x = 0 \) tells us that the Intermediate Value Theorem cannot predict values like \( f(c) = 0 \) in the given interval.