Question

Explorations Let \(f(x)=\left\\{\begin{array}{ll}{x^{2},} & {x \leq 1} \\ {2 x,} & {x>1}\end{array}\right.\) \begin{array}{ll}{\text { (a) Find } f^{\prime}(x) \text { for } x<1 .} & {\text { (b) Find } f^{\prime}(x) \text { for } x>1.2} \\ {\text { (c) Find } \lim _{x \rightarrow 1}-f^{\prime}(x) .2} &{\text { (d) Find } \lim _{x \rightarrow 1^{+}} f^{\prime}(x)}\end{array} \begin{array}{l}{\text { (e) Does } \lim _{x \rightarrow 1} f^{\prime}(x) \text { exist? Explain. }} \\ {\text { (f) Use the definition to find the left-hand derivative of } f^ {}} \\ {\text { at } x=1 \text { if it exists. } } \\ {\text { (g) Use the definition to find the right-hand derivative of } f} \\ {\text { at } x=1 \text { if it exists.}} \\ {\text { (h) Does \(f^{\prime}(1)\)} \text{exist?} \text{Explain.}} \end{array}

Short answer

a) 2, b) 2, c) 2, d) 2, e) Yes, f') does exist and equals 2, f) The left-hand derivative of f at x=1 is 2, g) The right-hand derivative of f at x=1 is 2, h) Yes, \(f'(1)\) exists and equals to 2.

Step-by-step solution

Differentiate the function

Differentiate both parts of the function separately, i.e., \(f'(x) = 2x\) for \(x \leq 1\) and \(f'(x) = 2\) for \(x > 1\).

Substitute for x

To find \(f'(x)\) for \(x < 1\), substitute x in \(\lim_{x \rightarrow 1}2x\), which gives the result of 2.

Substitute for x>1

To find \(f'(x)\) for \(x > 1\), substitute x in \(f'(x) = 2\), which is always 2 no matter the value of x.

Compute the limits

Next, calculate \(\lim_{x \rightarrow 1^-} f'(x)\) and \(\lim_{x \rightarrow 1^+} f'(x)\), this would result in 2 and 2 respectively.

Evaluate the existence of the derivative at x=1

Since the limits as x approaches 1 from both the left and the right are equal and finite, \(f'(x)\) exists at x=1 and is equal to the common value of the limits which is 2.

Check for the existence of the derivative

Use the definitions of left-hand and right-hand derivatives to confirm that \(f'(1)\) exists, and is equal to the calculated value: 2.

Key concepts

Limit of a Function

When studying calculus, one fundamental concept that often comes into play is the 'limit of a function'. This concept is pivotal to the understanding of continuity and differentiability. Simply put, the limit of a function as x approaches a given value is what the function's output (y-values) gets closer to as the input (x-values) gets closer to that specific point.

For piecewise functions like the one in our exercise, limits play a crucial role in determining at which points the function is continuous or has a derivative. In particular, the limit of the derivative of a piecewise function can indicate whether there is a 'kink' or 'corner' at a point, which is where the piecewise sections meet. In the given exercise, the limit of the derivative as x approaches 1 from both sides is the same (both are 2), hence, there's no sudden change in slope, meaning the piecewise function is smoothly connected at that point.

When the limits from the left-hand side and the right-hand side do not match, the function has a discontinuity at that point. In mathematical notation, we express this as \(\lim_{x \rightarrow a^-} f(x)\) for the left-hand limit and \(\lim_{x \rightarrow a^+} f(x)\) for the right-hand limit. If both these limits are equal, we can say the two-sided limit exists and is equal to their common value.

Left-Hand Derivative

The 'left-hand derivative' of a function at a certain point is the instantaneous rate of change of the function as you approach the point from the left side, or as x-values decrease towards the point. It is defined by the limit of the difference quotient as the x-values approach the target point from the left. For our function, the left-hand derivative at x=1 involves looking at the slope of \( x^2 \) as x approaches 1.

Formally, it's calculated as \( f'_{-}(a)=\lim_{h\rightarrow 0^-}\frac{f(a+h)-f(a)}{h}\). In essence, we are looking for how the function behaves just before it reaches the point of interest. The left-hand derivative tells us about the behavior of the function as it 'enters' the point. If we can find a finite value for this limit, then the left-hand derivative at that point exists.

Right-Hand Derivative

Conversely, the 'right-hand derivative' is all about understanding the behavior of the function as you approach the point from the right side, or as the x-values increase. It's symbolized by \( f'_{+}(a) \) and is defined similarly to the left-hand derivative, but with the limit as x approaches the target point from the right. For the function in the exercise, it would look at the slope of the line \( 2x \) for values just greater than 1.

Mathematically, it is given by \( f'_{+}(a)=\lim_{h\rightarrow 0^+}\frac{f(a+h)-f(a)}{h}\). The right-hand derivative tells us about the function's behavior as it 'exits' the point. It is a critical concept, especially for piecewise functions, that helps in understanding whether the function is smooth or has a sharp turn at a point.

Continuity of Derivative

The continuity of a derivative at a point is a condition that entails a derivative which does not have any abrupt changes in value. In other words, for a function to have a continuous derivative at a certain point, the limit of the derivative as it approaches the point from the left and the limit of the derivative as it approaches from the right must exist and be equal. Moreover, the derivative at the point itself must also exist and be equal to those limits.

In the context of the given exercise, we were asked whether the derivative at x=1 exists, which is a question about its continuity at that point. The answer turned out affirmative as the limits of the derivative from both sides at x=1 were equal. It's important for students to recognize that while a function might be continuous at a point, it doesn't necessarily imply that the derivative is continuous at that point. The piecewise function example helps illuminate this subtle distinction by demonstrating that even with a sharp corner (where the function is not differentiable), the function itself can still be continuous.