Problem 67 - Calculus Early Transcendentals

Question

You are asked in these exercises to determine whether a piecewise-defined function \(f\) is differentiable at a value \(x=x_{0}\) where \(f\) is defined by different formulas on different sides of \(x_{0} .\) You may use without proof the following result, which is a consequence of the Mean-Value Theorem (discussed in Section \(4.8) .\) Theorem. Let \(f\) be continuous at \(x_{0}\) and suppose that \(\lim _{x \rightarrow x_{0}} f^{\prime}(x)\) exists. Then \(f\) is differentiable at \(x_{0},\) and \(f^{\prime}\left(x_{0}\right)=\lim _{x \rightarrow x_{0}} f^{\prime}(x) .\) $$ \begin{array}{l}{\text { Let } \quad f(x)=\left\\{\begin{array}{ll}{x^{2},} & {x \leq 1} \\ {\sqrt{x},} & {x>1}\end{array}\right.} \\ {\text { Determine whether } f \text { is differentiable at } x=1 . \text { If so, find }} \\\ {\text { the value of the derivative there. }}\end{array} $$

Step-by-Step Solution

Answer

Function is not differentiable at \(x = 1\).

Step 1: Check Continuity at x=1

First, we check if the function is continuous at \(x = 1\). For continuity, the left-hand limit, right-hand limit, and the function value at \(x = 1\) must all be the same.\\[\lim_{{x \to 1^-}} f(x) = \lim_{{x \to 1^-}} x^2 = 1^2 = 1\] \\[\lim_{{x \to 1^+}} f(x) = \lim_{{x \to 1^+}} \sqrt{x} = \sqrt{1} = 1\] \\[f(1) = 1^2 = 1\] \Since the left-hand limit, right-hand limit, and the function value at \(x = 1\) are all equal, \(f\) is continuous at \(x = 1\).

Step 2: Find the Derivative for Each Piece

Calculate the derivative of each piece of the function. \For \(x \leq 1\), \(f(x) = x^2\), so \(f'(x) = 2x\). \For \(x > 1\), \(f(x) = \sqrt{x}\), so \(f'(x) = \frac{1}{2\sqrt{x}}\).

Step 3: Evaluate the Left-Hand Derivative at x=1

Find the left-hand derivative at \(x = 1\):\\[\lim_{{x \to 1^-}} f'(x) = \lim_{{x \to 1^-}} 2x = 2 \times 1 = 2\]

Step 4: Evaluate the Right-Hand Derivative at x=1

Find the right-hand derivative at \(x = 1\):\\[\lim_{{x \to 1^+}} f'(x) = \lim_{{x \to 1^+}} \frac{1}{2\sqrt{x}} = \frac{1}{2} \times \frac{1}{\sqrt{1}} = \frac{1}{2}\]

Step 5: Determine Differentiability at x=1

For \(f\) to be differentiable at \(x = 1\), both the left-hand and right-hand derivatives must be equal. Here, \(2 eq \frac{1}{2}\). Thus, the left-hand and right-hand derivatives at \(x = 1\) are not equal, meaning \(f\) is not differentiable at \(x = 1\).

Key Concepts

Piecewise Functions

Piecewise functions are a blend of multiple function expressions each defined over specific segments of the domain. Think of them like overlapping paths on a map, where each path has its own rule. This makes piecewise functions interesting and sometimes tricky when analyzing concepts like continuity and differentiability. It's like checking if you can smoothly transition from one path to another without any breaks.

To decide if a piecewise function is differentiable at a junction (a point where these 'paths' change), we use limits to examine the behavior near that point. We check the continuity at the junction first, ensuring smooth transitions without jumps in value. Then, we calculate derivatives of the functions on either side of the point and see if they match up.

• If both the left-hand and right-hand derivatives are equal at the junction, then the function is differentiable there.
• If they aren't equal, even if the function is continuous, it won't be differentiable at that point.

This concept is fundamental when identifying differentiation points on piecewise functions.

Mean Value Theorem

The Mean Value Theorem (MVT) is a powerful tool in calculus that provides a link between derivatives and the shape of a function. It essentially lays down a condition that between any two points on a smooth curve, there’s at least one point where the tangent's slope (derivative) matches the overall slope between the two points. It’s like saying if you run to a friend's house in the countryside, there’s some part of your journey where your instantaneous speed matches your average speed.

When applying the Mean Value Theorem to piecewise functions, especially around junctions, it helps determine if the function behaves nicely. This is important when calculating derivatives, as MVT often serves as a backdrop condition to ensure the limits for derivatives are valid. If a piecewise function is continuous and differentiable over an interval, any conclusion drawn using MVT holds weight. But remember:

• The function must be continuous on the closed interval.
• It has to be differentiable on the open interval.

In exercises, checking these conditions helps in determining if differentiability can be asserted at certain points using MVT.

Continuity

Continuity forms the basis of many calculus concepts, like differentiability. You can think of it like drawing a line on a piece of paper without lifting your pencil. For a function to be continuous at a particular point, it must not have any breaks, jumps, or holes at that point.

To check for continuity in piecewise functions, we follow a simple set of steps:

• Ensure that the limit of the function as it approaches from the left matches the limit from the right.
• Verify that these approaching limits also match the function's value directly at the point being studied.

If these conditions hold true, the function is seamless, hence continuous at that point.

In the given example, ensuring continuity first was crucial before assessing differentiability, as a differentiable function is always continuous (though a continuous function need not be differentiable – a subtle but important distinction!). If any of these continuity conditions fail, the piecewise function cannot be differentiable at that point, no matter the behavior of the individual segments.