Answer :
To determine the intervals on which the function [tex]\( f(x) = (x + 4)e^x \)[/tex] is increasing or decreasing and to find its local extrema, we follow these steps:
### 1. Find the first derivative of [tex]\( f(x) \)[/tex]:
The first step is to find the first derivative, [tex]\( f'(x) \)[/tex], which helps us analyze the intervals of increase or decrease.
Given:
[tex]\[ f(x) = (x + 4)e^x \][/tex]
Using the product rule, the first derivative is:
[tex]\[ f'(x) = \frac{d}{dx}[(x + 4)e^x] = (x + 4) \frac{d}{dx}[e^x] + e^x \frac{d}{dx}[x + 4] = (x + 4)e^x + e^x = (x + 5)e^x \][/tex]
### 2. Find the critical points:
The critical points occur where the first derivative is zero or undefined. The exponential function [tex]\( e^x \)[/tex] is never zero, so set the first derivative equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ (x + 5)e^x = 0 \][/tex]
[tex]\[ x + 5 = 0 \][/tex]
[tex]\[ x = -5 \][/tex]
So, the critical point is [tex]\( x = -5 \)[/tex].
### 3. Determine the second derivative for concavity and local extrema:
To identify whether the critical point is a local minimum or maximum, we use the second derivative, [tex]\( f''(x) \)[/tex]:
[tex]\[ f''(x) = \frac{d}{dx}[(x + 5)e^x] = (x + 5) \frac{d}{dx}[e^x] + e^x \frac{d}{dx}[x + 5] = (x + 5)e^x + e^x = (x + 6)e^x \][/tex]
Evaluate the second derivative at the critical point [tex]\( x = -5 \)[/tex]:
[tex]\[ f''(-5) = (-5 + 6)e^{-5} = 1 \cdot e^{-5} > 0 \][/tex]
Since [tex]\( f''(-5) > 0 \)[/tex], the function has a local minimum at [tex]\( x = -5 \)[/tex].
### 4. Intervals of increase and decrease:
To find intervals of increase and decrease, test the sign of the first derivative [tex]\( f'(x) = (x + 5)e^x \)[/tex] on either side of the critical point [tex]\( x = -5 \)[/tex]:
- For [tex]\( x < -5 \)[/tex], choose a test point like [tex]\( x = -6 \)[/tex]:
[tex]\[ f'(-6) = (-6 + 5)e^{-6} = -1 \cdot e^{-6} < 0 \][/tex]
Thus, [tex]\( f(x) \)[/tex] is decreasing in the interval [tex]\( (-\infty, -5) \)[/tex].
- For [tex]\( x > -5 \)[/tex], choose a test point like [tex]\( x = -4 \)[/tex]:
[tex]\[ f'(-4) = (-4 + 5)e^{-4} = 1 \cdot e^{-4} > 0 \][/tex]
Thus, [tex]\( f(x) \)[/tex] is increasing in the interval [tex]\( (-5, \infty) \)[/tex].
### Summary:
- Intervals of Increase: [tex]\( (-5, \infty) \)[/tex]
- Intervals of Decrease: [tex]\( (-\infty, -5) \)[/tex]
- Local Extrema: The function has a local minimum at [tex]\( x = -5 \)[/tex].
These conclusions come from analyzing the behavior of the first and second derivatives of [tex]\( f(x) = (x + 4)e^x \)[/tex].
### 1. Find the first derivative of [tex]\( f(x) \)[/tex]:
The first step is to find the first derivative, [tex]\( f'(x) \)[/tex], which helps us analyze the intervals of increase or decrease.
Given:
[tex]\[ f(x) = (x + 4)e^x \][/tex]
Using the product rule, the first derivative is:
[tex]\[ f'(x) = \frac{d}{dx}[(x + 4)e^x] = (x + 4) \frac{d}{dx}[e^x] + e^x \frac{d}{dx}[x + 4] = (x + 4)e^x + e^x = (x + 5)e^x \][/tex]
### 2. Find the critical points:
The critical points occur where the first derivative is zero or undefined. The exponential function [tex]\( e^x \)[/tex] is never zero, so set the first derivative equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ (x + 5)e^x = 0 \][/tex]
[tex]\[ x + 5 = 0 \][/tex]
[tex]\[ x = -5 \][/tex]
So, the critical point is [tex]\( x = -5 \)[/tex].
### 3. Determine the second derivative for concavity and local extrema:
To identify whether the critical point is a local minimum or maximum, we use the second derivative, [tex]\( f''(x) \)[/tex]:
[tex]\[ f''(x) = \frac{d}{dx}[(x + 5)e^x] = (x + 5) \frac{d}{dx}[e^x] + e^x \frac{d}{dx}[x + 5] = (x + 5)e^x + e^x = (x + 6)e^x \][/tex]
Evaluate the second derivative at the critical point [tex]\( x = -5 \)[/tex]:
[tex]\[ f''(-5) = (-5 + 6)e^{-5} = 1 \cdot e^{-5} > 0 \][/tex]
Since [tex]\( f''(-5) > 0 \)[/tex], the function has a local minimum at [tex]\( x = -5 \)[/tex].
### 4. Intervals of increase and decrease:
To find intervals of increase and decrease, test the sign of the first derivative [tex]\( f'(x) = (x + 5)e^x \)[/tex] on either side of the critical point [tex]\( x = -5 \)[/tex]:
- For [tex]\( x < -5 \)[/tex], choose a test point like [tex]\( x = -6 \)[/tex]:
[tex]\[ f'(-6) = (-6 + 5)e^{-6} = -1 \cdot e^{-6} < 0 \][/tex]
Thus, [tex]\( f(x) \)[/tex] is decreasing in the interval [tex]\( (-\infty, -5) \)[/tex].
- For [tex]\( x > -5 \)[/tex], choose a test point like [tex]\( x = -4 \)[/tex]:
[tex]\[ f'(-4) = (-4 + 5)e^{-4} = 1 \cdot e^{-4} > 0 \][/tex]
Thus, [tex]\( f(x) \)[/tex] is increasing in the interval [tex]\( (-5, \infty) \)[/tex].
### Summary:
- Intervals of Increase: [tex]\( (-5, \infty) \)[/tex]
- Intervals of Decrease: [tex]\( (-\infty, -5) \)[/tex]
- Local Extrema: The function has a local minimum at [tex]\( x = -5 \)[/tex].
These conclusions come from analyzing the behavior of the first and second derivatives of [tex]\( f(x) = (x + 4)e^x \)[/tex].
