Problem 104 - University Calculus: Early Transcendentals

Question

Suppose that \(f(x)=x^{2}\) and \(g(x)=|x| .\) Then the composites $$(f \circ g)(x)=|x|^{2}=x^{2} \quad \text { and } \quad(g \circ f)(x)=\left|x^{2}\right|=x^{2}$$ are both differentiable at \(x=0\) even though \(g\) itself is not differentiable at \(x=0 .\) Does this contradict the Chain Rule? Explain.

Step-by-Step Solution

Answer

No, it does not contradict the Chain Rule.

Step 1: Understanding Differentiability

The chain rule states that if two functions are differentiable, then their composite function is differentiable. For this exercise, we need to explore the differentiability of \(f \circ g\) and \(g \circ f\) at \(x = 0\).

Step 2: Examining the Functions f and g

The function \(f(x) = x^2\) is differentiable everywhere, including at \(x = 0\), with \(f'(x) = 2x\). The function \(g(x) = |x|\) is not differentiable at \(x = 0\) because its left-hand and right-hand derivatives do not match at this point.

Step 3: Calculating the Composite Function (f ∘ g)(x)

For the composite function \((f \circ g)(x) = f(g(x)) = |x|^2=x^2\), we note that \(x^2\) is the result, which is differentiable at all x, including \(x=0\). Thus, \((f \circ g)(x)\) is differentiable at \(x = 0\).

Step 4: Calculating the Composite Function (g ∘ f)(x)

For the composite function \((g \circ f)(x) = g(f(x)) = |x^2|=x^2\), the result is also \(x^2\). This function is differentiable everywhere, including at \(x = 0\). So, \((g \circ f)(x)\) is differentiable at \(x = 0\).

Step 5: Applying the Chain Rule

The Chain Rule requires both functions in the composition to be differentiable at the relevant point for the chain rule to hold. However, here the compositions \((f \circ g)(x)\) and \((g \circ f)(x)\) lead to the function \(x^2\), which is differentiable, despite \(g(x)\) not being differentiable at \(x=0\). This simply reflects that the chain rule does not apply because \(g(x)\) is not differentiable at \(x = 0\), but it does not prevent the compositions themselves from being differentiable.

Step 6: Conclusion

The differentiability of \((f \circ g)(x)\) and \((g \circ f)(x)\) does not contradict the Chain Rule because the rule is contingent on both functions being differentiable at the point in question. The fact that these compositions are differentiable is an exceptional case due to the particular functions involved, not a violation of mathematical principles.

Key Concepts

Function Composition

To better understand the exercise, let's start by exploring the concept of function composition. When we talk about function composition, we're simply referring to combining two functions in such a way that the output of one function becomes the input of another. For instance, if we have two functions, \( f(x) \) and \( g(x) \), we can create the composite functions \((f \circ g)(x)\) and \((g \circ f)(x)\). In our specific exercise,

• \((f \circ g)(x)\) means applying \( g \) first, then \( f \) to the result, giving us \( f(g(x)) \).
• \((g \circ f)(x)\) means applying \( f \) first, then \( g \) to the result, resulting in \( g(f(x)) \).

In other words, you're plugging one function into another. It can be a bit like mixing the ingredients of a recipe in a particular order to get the final dish—all the parts matter, but the order you mix them in really changes things! In practice, this process can help us tackle more complex functions by breaking them down and then building them back up through the rule of composition.

Differentiability

Now, what about differentiability? Differentiability is a concept that answers whether or not a function has a derivative at a certain point. It's very much tied to the idea of smoothness; if you imagine the graph of a function being smooth and without any sharp turns, it's likely differentiable there. In the context of our problem,

• The function \(f(x) = x^2\) is a classic example of a differentiable function, with \(f'(x) = 2x\), which clearly exists everywhere including at \(x = 0\).
• However, the function \(g(x) = |x|\) is not differentiable at \(x = 0\) because while approaching from the left you have \(-1\), and from the right \(1\); these don't match, creating a 'kink' or a sharp point in the graph.

The essence of examining differentiability is identifying whether functions possess a well-defined tangent at certain points. If the tangent exists (hint: ideally a straight line without abrupt changes), the function is differentiable. In the problem, although \(g(x)\) isn’t differentiable by itself at zero, the compositions are magically smooth and thus differentiable there.

Absolute Value Function

Finally, let's delve into the absolute value function, \(g(x) = |x|\), which plays a pivotal role in our exercise. The absolute value function basically takes any input, looks at its size without regard for its sign, and outputs that size as a positive number. This can be understood as a reflection process, where

• If \(x\) is positive or zero, \(g(x) = x\).
• If \(x\) is negative, \(g(x) = -x\).

The transition at \(x = 0\) is what makes \(g(x) = |x|\) non-differentiable at that point because it switches direction sharply. This results in a corner or a cusp at the origin of its graph. When we substitute \(|x|\) into compositions, like \((f \circ g)(x)\) or \((g \circ f)(x)\), the resulting function \(x^2\) is continuous and can be differentiated since those sharp turns disappear in the outcome. Thus, the end function being differentiable stems directly from these manipulations of the input through the absolute value operation while inherently smoothing out the graph's behavior.