Question

Find the sum function \((f+g)(x)\) if $$f(x)=\left\\{\begin{array}{ll}2 x+3 & \text { if } x<2 \\\x^{2}+5 x & \text { if } x \geq 2\end{array}\right.$$ and $$g(x)=\left\\{\begin{array}{ll}-4 x+1 & \text { if } x \leq 0 \\\x-7 & \text { if } x>0\end{array}\right.$$

Short answer

(f + g)(x) = -2x + 4 if x < 0, 3x - 4 if 0 ≤ x < 2, x^2 + 6x - 7 if x ≥ 2.

Step-by-step solution

Determine the intervals for the piecewise functions

Analyze the intervals for the functions given: - The function \( f(x) \) has different expressions for \( x < 2 \) and \( x \geq 2 \).- The function \( g(x) \) has different expressions for \( x \leq 0 \) and \( x > 0 \).

Identify combined intervals

To find the sum function, identify the combined intervals where each piecewise function applies:- For \( x < 0 \), use \( f(x) = 2x + 3 \) and \( g(x) = -4x + 1 \).- For \( 0 \leq x < 2 \), use \( f(x) = 2x + 3 \) and \( g(x) = x - 7 \).- For \( x \geq 2 \), use \( f(x) = x^2 + 5x \) and \( g(x) = x - 7 \).

Sum the functions in each interval

Add the corresponding expressions in each interval:1. For \( x < 0 \):\[ (f + g)(x) = (2x + 3) + (-4x + 1) = -2x + 4 \]2. For \( 0 \leq x < 2 \):\[ (f + g)(x) = (2x + 3) + (x - 7) = 3x - 4 \]3. For \( x \geq 2 \):\[ (f + g)(x) = (x^2 + 5x) + (x - 7) = x^2 + 6x - 7 \]

Write the final piecewise sum function

Combine the expressions obtained into one piecewise function:\[(f + g)(x) = \begin{cases}-2x + 4 & \text{if } x < 0 \3x - 4 & \text{if } 0 \leq x < 2 \x^2 + 6x - 7 & \text{if } x \, \geq 2\end{cases}\]

Key concepts

Sum of Functions

The sum of functions involves adding two or more functions together to create a new function. This process combines the outputs of the given functions for each input value. For example, if we have two functions, \(f(x)\) and \(g(x)\), their sum is represented as \((f + g)(x)\). This means we take the output of \(f(x)\) and add it to the output of \(g(x)\) for each input \(x\). Essentially, the new function \((f + g)(x)\) captures the combined effect of the original functions.

Piecewise-Defined Functions

Piecewise-defined functions are functions that have different expressions or formulas on different intervals of the input variable. These functions are often used to describe situations where a rule or relationship changes at certain points. For instance, a function might have one expression for input values less than 2 and a different expression for input values greater than or equal to 2. The notation for piecewise functions typically involves listing each expression along with its corresponding interval.

Interval Analysis

Interval analysis is the process of breaking down the domain of a function into specific intervals to understand and evaluate the behavior of the function within each interval. This is especially important when dealing with piecewise functions, as different expressions apply to different intervals. Analyzing intervals helps in identifying where each part of the piecewise function is applicable. Consider the function \(f(x)\) where it behaves differently for \(x<2\) and \(x\geq 2\). By separately evaluating \(x<0\), \(0 \text { }\leq \text { } x < 2\), and \(x \text { }\geq 2\), we can combine these behaviors to form a complete piecewise-defined function.

Adding Functions

Adding functions involves creating a new function where the value at any input is the sum of the values of the original functions at that same input. The procedure requires us to consider each interval of the piecewise-defined functions. For every interval, we sum the corresponding expressions. For example, given \(f(x)\) and \(g(x)\), to find \((f + g)(x)\) for \(x < 0\), combine the expressions for \(f(x)\) and \(g(x)\) in that specific interval: \((2x + 3) + (-4x + 1) = -2x + 4\). Repeating this for each interval helps us build the final piecewise sum function: \( (f + g)(x)\) as derived through step-by-step addition.