Question \( 3(4 \) goinst) Compare \( f(x)=x^{2}+4 \) and \( g(x)=-|x|+1 \) on the interval \( (-\infty, 0) \). Describe the functions as increasing or decreasing B I \( \underline{Y} \equiv \) S \( ^{*}, \approx \pi \Omega \)

To compare the functions \( f(x) = x^2 + 4 \) and \( g(x) = -|x| + 1 \) on the interval \( (-\infty, 0) \), we will analyze their behavior in this interval. ### Step 1: Analyze \( f(x) = x^2 + 4 \) 1. **Domain**: The function is defined for all \( x \). 2. **Behavior on \( (-\infty, 0) \)**: - The function \( f(x) \) is a parabola that opens upwards. - As \( x \) approaches \( -\infty \), \( f(x) \) increases because \( x^2 \) becomes very large. - At \( x = 0 \), \( f(0) = 0^2 + 4 = 4 \). - Therefore, on the interval \( (-\infty, 0) \), \( f(x) \) is **increasing**. ### Step 2: Analyze \( g(x) = -|x| + 1 \) 1. **Domain**: The function is defined for all \( x \). 2. **Behavior on \( (-\infty, 0) \)**: - For \( x < 0 \), \( |x| = -x \), so \( g(x) = -(-x) + 1 = x + 1 \). - This is a linear function with a slope of \( 1 \). - As \( x \) decreases (moves left), \( g(x) \) increases. - At \( x = 0 \), \( g(0) = 1 \). - Therefore, on the interval \( (-\infty, 0) \), \( g(x) \) is also **increasing**. ### Step 3: Compare \( f(x) \) and \( g(x) \) To compare the two functions, we can evaluate them at a few points in the interval \( (-\infty, 0) \): - At \( x = -1 \): - \( f(-1) = (-1)^2 + 4 = 1 + 4 = 5 \) - \( g(-1) = -|-1| + 1 = -1 + 1 = 0 \) - At \( x = -2 \): - \( f(-2) = (-2)^2 + 4 = 4 + 4 = 8 \) - \( g(-2) = -|-2| + 1 = -2 + 1 = -1 \) From these evaluations, we can see that: - \( f(-1) = 5 > g(-1) = 0 \) - \( f(-2) = 8 > g(-2) = -1 \) ### Conclusion - Both functions \( f(x) \) and \( g(x) \) are **increasing** on the interval \( (-\infty, 0) \). - However, \( f(x) \) is always greater than \( g(x) \) in this interval. Thus, we can summarize: - \( f(x) \) is increasing and greater than \( g(x) \) on \( (-\infty, 0) \). - \( g(x) \) is also increasing but remains below \( f(x) \) in this interval.