Problem 37 - The Calculus with Analytic Geometry

Question

Find formulas for \((f \circ g)(x)\) if $$ f(x)=\left\\{\begin{array}{ll} 0 & \text { if } x<0 \\ x^{2} & \text { if } 0 \leq x \leq 1 \\ 0 & \text { if } x>1 \end{array} \text { and } g(x)= \begin{cases}1 & \text { if } x<0 \\ 2 x & \text { if } 0 \leq x \leq 1 \\ 1 & \text { if } x>1\end{cases}\right. $$

Step-by-Step Solution

Answer

\( (f \circ g)(x) = \begin{cases} 1 & \text{if } x < 0 \ 4x^2 & \text{if } 0 \leq x \leq 1 \ 1 & \text{if } x > 1 \end{cases} \)

Step 1: Understand the functions

First, understand the piecewise functions for both \(f(x)\) and \(g(x)\). For \(f(x)\): \[f(x) = \begin{cases} 0 & \text{if } x < 0 \ x^2 & \text{if } 0 \leq x \leq 1 \ 0 & \text{if } x > 1 \end{cases}\]For \(g(x)\): \[ g(x) = \begin{cases} 1 & \text{if } x < 0 \ 2x & \text{if } 0 \leq x \leq 1 \ 1 & \text{if } x > 1 \end{cases}\]

Step 2: Define the composition \((f \circ g)(x)\)

Recall that the composition \((f \circ g)(x)\) is defined as \(f(g(x))\). To determine this, you need to substitute \(g(x)\) into \(f(x)\).

Step 3: Determine \(f(g(x))\) for different intervals

Evaluate \(f(g(x))\) based on the intervals defined by \(g(x)\): - For \(x < 0\): \(g(x)=1\), so \(f(g(x)) = f(1)\) and since \(0 \le 1 \le 1\), \(f(1) = 1^2 = 1\).- For \(0 \le x \le 1\): \(g(x)=2x\), so \(f(g(x)) = f(2x)\). Consider the ranges of \(2x\). If \(0 \le 2x \le 1\), \(f(2x) = (2x)^2 = 4x^2\).- For \(x > 1\): \(g(x)=1\), so \(f(g(x)) = f(1)\). As deduced earlier, \(f(1) = 1\).

Step 4: Combine the results

Combine the results from the different intervals: - For \(x < 0\), \((f \circ g)(x) = 1\).- For \(0 \le x \le 1\), \((f \circ g)(x) = 4x^2\).- For \(x > 1\), \((f \circ g)(x) = 1\).

Key Concepts

Piecewise Functions

A piecewise function is defined by different expressions depending on the interval of the input value. Essentially, it 'pieces' together different parts of functions to create a single, multi-part function. In our given exercise, we examine two piecewise functions, \( f(x) \) and \( g(x) \).

For \( f(x) \), the expression changes based on the value of x:
\[ f(x) = \begin{cases} 0 & \text{if } x < 0 \ x^2 & \text{if } 0 \leq x \leq 1 \ 0 & \text{if } x > 1 \end{cases} \]
Similarly, \( g(x) \) changes based on three intervals:
\[ g(x) = \begin{cases} 1 & \text{if } x < 0 \ 2x & \text{if } 0 \leq x \leq 1 \ 1 & \text{if } x > 1 \end{cases} \]
The key point is understanding which expression applies to the given input value of x. This approach allows greater flexibility in modeling situations where behavior changes over different ranges.

Function Evaluation

Function evaluation involves determining the output of a function when supplied with a given input. In this exercise, we need to evaluate the composition of functions, \( (f \circ g)(x) \), which means applying \( g(x) \) first and then using that result as an input to \( f(x) \).

Here's a step-by-step breakdown:

• For \( x < 0 \): \( g(x) = 1 \), so \( f(g(x)) = f(1) \). Since 1 fits the middle interval of \( f(x) \), we compute \( f(1) = 1^2 = 1 \).


• For \( 0 \leq x \leq 1 \): \( g(x) = 2x \), so \( f(g(x)) = f(2x) \). Since 2x falls within the middle interval of \( f(x) \) when \( 0 \leq 2x \leq 1 \) (i.e., \( 0 \leq x \leq 0.5 \)), we find \( f(2x) = (2x)^2 = 4x^2 \).


• For \( x > 1 \): \( g(x) = 1 \), so \( f(g(x)) = f(1) \). As calculated earlier, \( f(1) = 1 \).

By evaluating the function step by step, we ensure accuracy and understand how the composition behaves over different intervals.

Interval Analysis

Interval analysis is crucial in understanding piecewise functions and their compositions. It helps us determine which part of the function applies to a specific input.

In this exercise, we segmented the input values into three intervals:

• For \( x < 0 \), \( g(x) = 1 \), and thus \( f(1) = 1 \).


• For \( 0 \leq x \leq 1 \), \( g(x) = 2x \). Within this interval, 2x falls into the range \( 0 \leq 2x \leq 1 \), and we compute \( f(2x) = 4x^2 \).


• For \( x > 1 \), \( g(x) = 1 \), and so \( f(1) = 1 \).

Combining these, we establish the final composite function:

\[ (f \circ g)(x) = \begin{cases} 1 & \text{if } x < 0 \ 4x^2 & \text{if } 0 \leq x \leq 1 \ 1 & \text{if } x > 1 \end{cases} \]
Analyzing intervals helps in breaking down complex functions into manageable parts, making it easier to understand their behavior thoroughly.