Problem 42 - Calculus Early Transcendentals

Question

(a) Let $$ f(x)=\left\\{\begin{array}{ll}{x^{2},} & {x \leq 0} \\ {x^{2}+1,} & {x>0}\end{array}\right. $$ Show that $$ \lim _{x \rightarrow 0^{-}} f^{\prime}(x)=\lim _{x \rightarrow 0^{+}} f^{\prime}(x) $$ but that \(f^{\prime}(0)\) does not exist. (b) Let $$ f(x)=\left\\{\begin{array}{ll}{x^{2},} & {x \leq 0} \\ {x^{3},} & {x>0}\end{array}\right. $$ Show that \(f^{\prime}(0)\) exists but \(f^{\prime \prime}(0)\) does not.

Step-by-Step Solution

Answer

Part (a) Derivatives match on both sides; \( f'(0) \) does not exist. Part (b) \( f'(0) \) exists; \( f''(0) \) does not.

Step 1: Describe the Limits of the Derivatives for Each Piece

First, find the derivative of each piece of the function \( f(x) \). For \( x \leq 0 \), the derivative is \( f'(x) = 2x \). For \( x > 0 \), the derivative is \( f'(x) = 2x \) as well (derivative of \( x^2 + 1 \)). Evaluate the limits \( \lim_{x \to 0^-} f'(x) \) and \( \lim_{x \to 0^+} f'(x) \). Both are equal to \( 0 \).

Step 2: Verify Non-existence of the Derivative at Zero

Check if \( f'(0) \) exists by evaluating the limit \( \lim_{h \to 0} \frac{f(h) - f(0)}{h} \). Since for \( h>0 \), \( f(h) = h^2 + 1 \) and for \( h \leq 0 \), \( f(h) = h^2 \), the limit behaves differently from each side, showing \( f'(0) \) does not exist.

Step 3: Evaluate the Derivative of the Second Function

Find the derivative of the second function \( f(x) = \left\{x^2, x \leq 0; x^3, x > 0\right\} \). For \( x \leq 0 \), the derivative is \( 2x \) and for \( x > 0 \), the derivative is \( 3x^2 \).\\ Evaluate \( \lim_{x \to 0^-} f'(x) \) and \( \lim_{x \to 0^+} f'(x) \). Both limits yield \( 0 \).

Step 4: Show that the Derivative at Zero Exists

Calculate \( f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} \). Substitute \( f(h) = h^2 \) for \( h \leq 0 \) and \( f(h) = h^3 \) for \( h > 0 \). Both sides give the limit as \( 0 \), thus \( f'(0) = 0 \) exists.

Step 5: Examine the Second Derivative Existence

For \( x \leq 0 \), the second derivative \( f''(x) = 2 \). For \( x > 0 \), \( f''(x) = 6x \). Evaluate \( \lim_{x \to 0^-} f''(x) \) and \( \lim_{x \to 0^+} f''(x) \). These limits are not equal (2 and 0, respectively), showing the second derivative at zero does not exist.

Key Concepts

Derivatives

In calculus, derivatives measure how a function changes as its input changes. If you have a function like our example with piecewise components, each segment of the function can have its own derivative. Derivatives help us understand the rate of change or the slope of the graph at any given point.

• For the function \(f(x) = x^2\), the derivative is \(f'(x) = 2x\). This tells us the slope of the tangent line to \(f(x)\) at any point \(x\).
• In our example function, for the interval where \(x \leq 0\), both pieces of the function have a derivative of \(2x\), meaning the rate of change is the same linear function in both pieces for that domain.

The key to understanding the role of derivatives is to break the function into manageable parts and find the derivative of each part separately.

Non-existence of Derivative

A derivative at a particular point, like \(f'(0)\), may not always exist, even if the limits of the derivatives from either side are the same. In practice, for \(f'(0)\) to exist, the function must have the same slope approaching zero from both the left and the right. When you calculate the derivative using:\[f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h}\]You'll consider:

• For \(h > 0\), \(f(h) = h^2 + 1\) diverges from \(f(0) = 0^2 = 0\) because it includes a constant shift of +1, leading to diverging approaches.
• As \(h \leq 0\), \(f(h) = h^2\) aligns smoothly with \(f(0)\), following the same curve with no extra offset.

Since these are unequal, \(f'(0)\) does not exist for \(f(x) = x^2 + 1\) at zero. Observing where these differ helps understand the reasons a derivative might be undefined at a point.

Piecewise Functions

Piecewise functions are defined by different expressions over various segments of their domain. Understanding these is crucial because each of those segments can behave differently in terms of limits and derivatives.

• For our function \(f(x)\), it’s piecewise because it has different definitions for \(x \leq 0\) and \(x > 0\).
• This change can affect both the existence and the behavior of its derivatives at transition points (like \(x = 0\)). Understanding whether or not a derivative exists at such a point often involves checking if the limits of the derivatives approach the same value from both sides.

In more complex piecewise functions, each piece needs to be evaluated for continuity and differentiability separately, ensuring that all transitions between pieces are smooth wherever necessary for derivatives to exist.