Problem 82 - Calculus: Early Transcendentals

Question

Let \( f(x) = \left\\{ \begin{array}{ll} 0 & \mbox{if \) x < 0 \(}\\\ x & \mbox{if \) 0 \le x \le 1 \(}\\\ 2 - x & \mbox{if \) 1 < x \le 2 \(}\\\ 0 & \mbox{if \) x > 2 \(} \end{array} \right.\) and $$ g(x) = \int^x_0 f(t) \, dt $$ (a) Find an expression for \( g(x) \) similar to the one for \( f(x) \). (b) Sketch the graphs of \( f \) and \( g \). (c) Where is \( f \) differentiable? Where is \( g \) differentiable?

Step-by-Step Solution

Answer

(a) \( g(x) = 0 \) if \( x < 0 \); \( g(x) = \frac{x^2}{2} \) if \( 0 \leq x \leq 1 \); \( g(x) = 2x - \frac{x^2}{2} \) if \( 1 < x \le 2 \); \( g(x) = 2 \) if \( x > 2 \). (b) Graphs vary with intervals. (c) \( f \) is differentiable on \( (0, 1) \), \( (1, 2) \); \( g \) on \( (0, 2) \).

Step 1: Understanding the Function f(x)

The function \( f(x) \) is a piecewise function divided into four intervals:- \( f(x) = 0 \) if \( x < 0 \)- \( f(x) = x \) if \( 0 \leq x \leq 1 \)- \( f(x) = 2 - x \) if \( 1 < x \leq 2 \)- \( f(x) = 0 \) if \( x > 2 \).

Step 2: Analyzing the Function g(x)

The function \( g(x) = \int_0^x f(t) \, dt \) is the integral of \( f(t) \) from \( 0 \) to \( x \). You must consider the intervals of \( f(x) \) when finding \( g(x) \), since \( f(x) \) is piecewise.

Step 3: Calculating g(x) for Each Interval

- **If \( x < 0 \):** \( g(x) = 0 \) since the limits of integration are invalid (upper limit is less than the lower limit).- **If \( 0 \leq x \leq 1 \):** Compute \( g(x) = \int_0^x t \, dt = \frac{x^2}{2} \).- **If \( 1 < x \leq 2 \):** Compute \( g(x) = \int_0^1 t \, dt + \int_1^x (2-t) \, dt = \left[\frac{t^2}{2}\right]_0^1 + \left[2t - \frac{t^2}{2}\right]_1^x = \frac{1}{2} + 2(x - 1) - \frac{x^2 - 1}{2} \).- **If \( x > 2 \):** \( g(x) \) is constant since \( f(x) = 0 \); thus, \( g(x) = \int_0^2 f(t) \, dt \).

Step 4: Simplifying g(x) by Interval

For interval \( 1 < x \le 2 \), further simplify:\[g(x) = \frac{1}{2} + 2(x - 1) - \frac{x^2 - 1}{2} = \frac{3}{2} + 2x - 2 - \frac{x^2}{2} + \frac{1}{2}\]Returns to:\[g(x) = 2x - \frac{x^2}{2}\]At \( x = 2 \), solve:\[g(2) = 2(2) - \frac{4}{2} = 4 - 2 = 2\]

Step 5: Sketching Graphs

Graph \( f(x) \): a zero line, increasing line \( 0 \leq x \leq 1 \), and a decreasing line \( 1 < x \leq 2 \).Graph \( g(x) \): convex up from \( 0 \) to \( 1 \), more accelerated convex from \( 1 \) to \( 2 \), zeroed out afterwards.

Step 6: Analyzing Differentiability

The function \( f(x) \) is differentiable at \( x \) in the intervals \( (0, 1) \) and \( (1, 2) \), but not at the points where the definition changes.The function \( g(x) \), being an integral of a piecewise continuous function \( f(x) \), is differentiable over the entire domain \( (0, 2) \).

Key Concepts

Piecewise Functions

A piecewise function is a function composed of different expressions over various intervals of its domain. It allows for different behaviors in different sections of its input values, making it flexible for situations where a single expression might not suffice.
In our example, the function \( f(x) \) is given as piecewise with the following segments:

• \( f(x) = 0 \) for \( x < 0 \) and \( x > 2 \).
• \( f(x) = x \) between \( 0 \leq x \leq 1 \), showing a simple linear increase.
• \( f(x) = 2 - x \) from \( 1 < x \leq 2 \), demonstrating a linear decrease.

These segments mean that the behavior of \( f(x) \) changes at the points \( x = 0 \), \( x = 1 \), and \( x = 2 \). Understanding how each segment works and connects at these points helps to grasp the full behavior of piecewise functions. This function is particularly useful for constructing models that involve constraints or specific conditions over certain ranges.

Differentiability

Differentiability refers to the ability of a function to have a derivative at a particular point. For a function to be differentiable at a point, the function must be continuous there, and its slope should not abruptly change. In piecewise functions, differentiability often varies across its different segments.
For \( f(x) \), differentiability is evaluated in the following manner:

• It is differentiable in the open intervals \( (0, 1) \) and \( (1, 2) \) where the respective expressions \( f(x) = x \) and \( f(x) = 2-x \) are linear.
• It is not differentiable at the points where the definition changes: \( x = 0 \), \( x = 1 \), and \( x = 2 \), because there are abrupt changes in direction or slope.

On the other hand, because \( g(x) \) is obtained by integrating \( f(x) \), it results in a function that is smoothly connected across intervals, making it differentiable over the entire domain \( (0, 2) \). This is often true for integrals of piecewise continuous functions provided the pieces of the original function connect smoothly.

Graph Sketching

Sketching the graphs of piecewise functions and their integral can reveal much about their behavior. Let's break down the process for \( f(x) \) and \( g(x) \), step-by-step:

• **\( f(x) \):** Start with a horizontal line at zero for \( x < 0 \) and \( x > 2 \). From \( 0 \leq x \leq 1 \), draw a line increasing linearly to \( x = 1 \). Extend a linear decreasing line from \( x = 1 \) to \( x = 2 \). This shows a simple linear change within each interval.
• **\( g(x) \):** Understanding that \( g(x) \), as the integral, can be constructed by accumulating area. It starts from zero; \( g(x) = \frac{x^2}{2} \) as a parabola from \( 0 \leq x \leq 1 \). From \( 1 < x \leq 2 \), \( g(x) = 2x - \frac{x^2}{2} \) indicates a more rapid increase until it tops out at a value of \( 2 \) at \( x = 2 \).

Graphing helps to visually assess where changes and breaks occur, particularly in examining points of differentiability and behavior across intervals. Sketching serves as a powerful tool to interpret the overall nature of a function within a problem.