Problem 68 - Calculus Early Transcendentals: Pearson New International Edition

Question

The symmetric derivative \(f_{s}(x)\) is defined by $$f_{s}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x-h)}{2 h}$$ Show that if \(f^{\prime}(x)\) exists then \(f_{s}(x)\) exists, but that the converse is false.

Step-by-Step Solution

Answer

If a standard derivative exists, the symmetric derivative must exist. The converse is false; consider the function \(f(x) = |x|\).

Step 1: Understand symmetric derivative

The symmetric derivative \(f_{s}(x)\) of a function \(f(x)\) is given by \(f_{s}(x) = \lim _{h \rightarrow 0} \frac{f(x+h) - f(x-h)}{2h}\). This definition is similar to the standard derivative but uses a symmetric difference quotient.

Step 2: Show existence of symmetric derivative when standard derivative exists

Assume \(f^{\prime}(x)\) exists, meaning \(\lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = f'(x)\). For the symmetric derivative, consider:\[\frac{f(x+h) - f(x-h)}{2h} = \frac{(f(x+h) - f(x)) + (f(x) - f(x-h))}{2h}.\] This expression is the average of the forward and backward difference quotients. As \(h \to 0\), both \(f(x+h) - f(x)\) and \(f(x) - f(x-h)\) approach 0, leading to:\[\lim_{h \to 0} \frac{f(x+h) - f(x-h)}{2h} = \frac{f'(x) + f'(x)}{2} = f'(x).\]Thus, if \(f^{\prime}(x)\) exists, then \(f_{s}(x)\) also exists, and \(f_{s}(x) = f'(x).\)

Step 3: Show converse is false using counterexample

To show the converse is false, we need a function with an existing symmetric derivative but no standard derivative. Consider the function \(f(x) = |x|\), which is not differentiable at \(x = 0\) since the left-hand and right-hand limits do not match:- Right-hand limit: \(\lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{|h|}{h} = 1.\)- Left-hand limit: \(\lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{|h|}{h} = -1.\)The symmetric derivative evaluates to:\[f_{s}(0) = \lim_{h \to 0} \frac{f(h) - f(-h)}{2h} = \lim_{h \to 0} \frac{|h| - |h|}{2h} = \lim_{h \to 0} \frac{0}{2h} = 0.\]Therefore, \(f_{s}(0)\) exists even though \(f^{\prime}(0)\) does not.

Key Concepts

Existence of Derivative

The existence of a derivative at a point for a function is a fundamental concept in calculus. It signifies whether we can find a tangent line at that point. More formally, for a function \( f(x) \), the derivative at a point \( x \), denoted as \( f'(x) \), exists if the limit \( \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \) converges to a fixed value.

• If this limit exists, \( f(x) \) is differentiable at that point.
• The value represents the slope of the tangent at \( x \).

Understanding when a derivative exists helps us describe smooth behavior in functions.

Standard Derivative

The standard derivative is often what first comes to mind when we think about derivatives. It measures how a function changes as its input changes, essentially describing the limit of the function’s instantaneous rate of change. Given a function \( f(x) \), the standard derivative \( f'(x) \) is defined by:\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

• The numerator \( f(x+h)-f(x) \) indicates the change in the function's value.
• The denominator \( h \) shows the change in input.

In practical terms, calculating this derivative gives insight into the behavior and properties of the function at specific points.

Differentiability

Differentiability is a concept that goes a step further from the mere existence of derivatives. A function is said to be differentiable at a point if its derivative exists at that point. It is important to note:

• If \( f'(x) \) exists, the function is smooth or continuous at \( x \).
• Differentiability implies continuity but not vice-versa.

Differentiability across an interval implies that not only does the derivative exist at every point in the interval, but also that the function has no abrupt changes, jumps, or corners within that interval.

Counterexample

A counterexample provides a concrete case where a general statement fails. In our context, it helps demonstrate that even if a symmetric derivative exists, the standard derivative might not.
The function \( f(x) = |x| \) serves as a classic counterexample:

• At \( x=0 \), the standard derivative does not exist because the limit \( \lim_{h \to 0} \frac{f(h) - f(0)}{h} \) does not converge due to differing left and right-hand limits.
• However, the symmetric derivative \( f_s(0) \) exists because \( \lim_{h \to 0} \frac{f(h) - f(-h)}{2h} = 0 \), showing equality in approaching zero from both sides symmetrically.

This example illustrates when conventional intuitions about derivatives may not fully apply, offering insight into the subtleties of calculus.