Explore Westonci.ca, the premier Q&A site that helps you find precise answers to your questions, no matter the topic. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
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].
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.
