Problem 57 - Single Variable Calculus

Question

In Exercises 57 and 58 , let \(f(x)=x\left(1-x^{2}\right)\), and let \(g\) be the signum (or sign) function defined by $$ g(x)=\left\\{\begin{array}{ll} -1 & \text { if } x<0 \\ 0 & \text { if } x=0 \\ 1 & \text { if } x>0 \end{array}\right. $$ Show that \(f \circ g\) is continuous on \((-\infty, \infty)\). Does this contradict Theorem 6 ?

Step-by-Step Solution

Answer

In this problem, we found that the composition function \(f(g(x)) = 0\) for all \(x\). Since it is a constant function, it is continuous on \((-\infty, \infty)\). This doesn't contradict Theorem 6, which states a condition for continuity but does not provide an exhaustive list of possible cases for continuous composition functions. Thus, the composition function \(f \circ g\) is continuous everywhere despite the discontinuity of \(g(x)\) at \(x=0\).

Step 1: Find \(f(g(x))\)

First, we substitute \(g(x)\) into \(f(x)\), i.e., find \(f(g(x))\). Since we have three cases for the signum function: 1. \(g(x) = -1\) if \(x < 0\) 2. \(g(x) = 0\) if \(x = 0\) 3. \(g(x) = 1\) if \(x > 0\) We calculate \(f(g(x))\) for each case:

Step 2: Case 1: \(x < 0\)

For \(x < 0\), we have \(g(x) = -1\). So, \(f(g(x)) = f(-1)\). Now, substituting \(-1\) into \(f(x) = x(1 - x^2)\): $$ f(g(x)) = f(-1) = (-1)(1 - (-1)^2) = -1(1 - 1) = 0 $$

Step 3: Case 2: \(x = 0\)

For \(x = 0\), we have \(g(x) = 0\). So, \(f(g(x)) = f(0)\). Now, substituting \(0\) into \(f(x) = x(1 - x^2)\): $$ f(g(x)) = f(0) = (0)(1 - 0^2) = 0 $$

Step 4: Case 3: \(x > 0\)

For \(x > 0\), we have \(g(x) = 1\). So, \(f(g(x)) = f(1)\). Now, substituting \(1\) into \(f(x) = x(1 - x^2)\): $$ f(g(x)) = f(1) = (1)(1 - 1^2) = 1(1 - 1) = 0 $$ As a result, for all cases, \(f(g(x)) = 0.\)

Step 2: Check continuity

Now, we need to check the continuity of \(f(g(x))\) on \((-\infty, \infty)\). Since \(f(g(x))\) is a constant function equal to \(0\), it is continuous on \((-\infty, \infty)\).

Step 3: Discuss the relation to Theorem 6

Theorem 6 states that if \(g\) is continuous at \(f(a)\) and \(f\) is continuous at \(a\), then the composition function \(f \circ g\) is continuous at \(a\). However, the given exercise shows that \(f \circ g\) is continuous everywhere even though \(g(x)\) (the signum function) is not continuous at \(x=0\). This doesn't contradict Theorem 6 because it states a condition when the composition function will be continuous but does not provide the exhaustive list of possible cases that could lead to a continuous composition function. In this case, thecomposition function is indeed continuous everywhere despite the discontinuity of \(g(x)\) at \(x=0\).

Key Concepts

Signum Function

The signum function, also known as the sign function, is an important mathematical function that is used to extract the sign of a real number. It is defined as follows for any real number 'x':

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Continuous Function

Understanding continuity is fundamental in calculus. A function is continuous at a point when there is no interruption in the graph of the function at that point. In layman's terms, if you can draw the function from start to end without lifting your pen from the paper, the function is continuous.For a more formal definition, a function 'f' is continuous at a point 'a' if the following three conditions are met:

• The function 'f(a)' is defined.
• The limit of 'f(x)' as 'x' approaches 'a' exists.
• The limit of 'f(x)' as 'x' approaches 'a' equals 'f(a)'.

This condition must be true for every point in the function's domain for it to be called continuous over that range. Since real-life situations are often modeled using functions that behave without abrupt changes, continuous functions are widely applicable in science, engineering, and economics.

Theorem of Continuous Functions

When working with compositions of functions, it's valuable to know when the resulting function will be continuous. This is outlined in a prominent result in calculus known as the theorem of continuous functions or sometimes more specifically, as the chain rule for limits. Essentially, it states that if function 'g' is continuous at 'a', and function 'f' is continuous at 'g(a)', then their composition, expressed as 'f(g(x))', is also continuous at 'a'.

Applying the Theorem to Compositions

This principle is powerful because it allows us to predict the behavior of complex functions built from simpler ones. However, it's important to note that this theorem provides a sufficient condition for continuity of compositions, not a necessary one. This means while the theorem tells us one way that a composition can be continuous, there are other ways in which compositions might maintain continuity, even if one of the functions is not continuous everywhere. For example, if the inner function leads to a constant output, as shown in the original exercise with the signum and polynomial functions, the composition will be continuous regardless of the discontinuities of the inner function.Understanding this theorem is crucial for correctly applying concepts in advanced calculus, and it provides a solid foundation for grasping the nuances of function behavior.