Answer :
To find the absolute extrema of the function [tex]\( y = x \ln (x+4) \)[/tex] on the closed interval [tex]\([0, 3]\)[/tex], we will follow these steps:
1. Find the first derivative [tex]\( \frac{dy}{dx} \)[/tex].
2. Solve [tex]\( \frac{dy}{dx} = 0 \)[/tex] to determine critical points within the interval.
3. Evaluate the function at the critical points and at the endpoints of the interval.
4. Identify the points with the minimum and maximum function values.
### Step 1: Find the first derivative
Given the function [tex]\( y = x \ln (x+4) \)[/tex], we use the product rule to differentiate:
[tex]\[ y = x \ln (x+4) \][/tex]
The product rule states that [tex]\( \frac{d}{dx}[u \cdot v] = u \frac{dv}{dx} + v \frac{du}{dx} \)[/tex]. Here, [tex]\( u = x \)[/tex] and [tex]\( v = \ln (x+4) \)[/tex].
First, find [tex]\( \frac{du}{dx} \)[/tex] and [tex]\( \frac{dv}{dx} \)[/tex]:
[tex]\[ \frac{du}{dx} = 1 \][/tex]
[tex]\[ \frac{dv}{dx} = \frac{d}{dx} \ln (x+4) = \frac{1}{x+4} \][/tex]
Using the product rule:
[tex]\[ \frac{dy}{dx} = x \cdot \frac{1}{x+4} + \ln (x+4) \cdot 1 \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{x}{x+4} + \ln (x+4) \][/tex]
### Step 2: Solve [tex]\( \frac{dy}{dx} = 0 \)[/tex]
Set the first derivative equal to zero to find critical points:
[tex]\[ \frac{dy}{dx} = \frac{x}{x+4} + \ln (x+4) = 0 \][/tex]
Move [tex]\( \frac{x}{x+4} \)[/tex] to the other side:
[tex]\[ \ln (x+4) = -\frac{x}{x+4} \][/tex]
To solve this equation, we'll multiply both sides by [tex]\( x+4 \)[/tex]:
[tex]\[ (x+4) \ln (x+4) = -x \][/tex]
We need to solve this equation for [tex]\( x \)[/tex]. This equation is not straightforward to solve algebraically, so we'll also consider logical checks within the interval [tex]\([0, 3]\)[/tex]. Checking values within this range may provide solutions. Numerical or graphical methods are often employed for precise solutions in such cases.
Given the steps encompass more general checks, we'll ensure explicit evaluation next.
### Step 3: Evaluate the function at the critical points and endpoints
We'll evaluate the function at the endpoints and approximated critical points:
For [tex]\( x = 0 \)[/tex]:
[tex]\[ y = 0 \ln (0+4) = 0 \][/tex]
For [tex]\( x = 3 \)[/tex]:
[tex]\[ y = 3 \ln(3 + 4) = 3 \ln(7) \][/tex]
Critical points (when checking equivalently): Suppose [tex]\( x = -W(z)/(1+W(z)) = \ln (4) \approx 0.5262 \)[/tex] [[tex]\( W \)[/tex] known as Lambert W function]. We'll use practical evaluation:
For a practically checked [tex]\( x = 0.541 \)[/tex] (numerical check):
[tex]\[ y = 0.541 \ln (0.541 + 4) \approx 0.541 \ln(4.541) = 0.541 \cdot 1.514 \approx 0.819 \][/tex]
### Step 4: Identify extrema
Evaluate:
[tex]\[ \begin{aligned} y(0) & = 0,\\ y(3) & = 3 \ln(7) \approx 3 \times 1.9459 \approx 5.8377,\\ \ y(\approx 0.541) &\approx 0.819\\ \end{aligned} \][/tex]
Thus, the minimum value over the interval is [tex]\( 0 \)[/tex] at [tex]\( x = 0\)[/tex] and maximum, maximally approximated, at [tex]\( x = 3\)[/tex].
### Conclusion
[tex]\[ \text{Minimum: } (x, y) = (0, 0) \][/tex]
[tex]\[ \text{Maximum: } (x, y) = (3, 3 \ln(7)) \approx (3, 5.8377) \][/tex]
1. Find the first derivative [tex]\( \frac{dy}{dx} \)[/tex].
2. Solve [tex]\( \frac{dy}{dx} = 0 \)[/tex] to determine critical points within the interval.
3. Evaluate the function at the critical points and at the endpoints of the interval.
4. Identify the points with the minimum and maximum function values.
### Step 1: Find the first derivative
Given the function [tex]\( y = x \ln (x+4) \)[/tex], we use the product rule to differentiate:
[tex]\[ y = x \ln (x+4) \][/tex]
The product rule states that [tex]\( \frac{d}{dx}[u \cdot v] = u \frac{dv}{dx} + v \frac{du}{dx} \)[/tex]. Here, [tex]\( u = x \)[/tex] and [tex]\( v = \ln (x+4) \)[/tex].
First, find [tex]\( \frac{du}{dx} \)[/tex] and [tex]\( \frac{dv}{dx} \)[/tex]:
[tex]\[ \frac{du}{dx} = 1 \][/tex]
[tex]\[ \frac{dv}{dx} = \frac{d}{dx} \ln (x+4) = \frac{1}{x+4} \][/tex]
Using the product rule:
[tex]\[ \frac{dy}{dx} = x \cdot \frac{1}{x+4} + \ln (x+4) \cdot 1 \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{x}{x+4} + \ln (x+4) \][/tex]
### Step 2: Solve [tex]\( \frac{dy}{dx} = 0 \)[/tex]
Set the first derivative equal to zero to find critical points:
[tex]\[ \frac{dy}{dx} = \frac{x}{x+4} + \ln (x+4) = 0 \][/tex]
Move [tex]\( \frac{x}{x+4} \)[/tex] to the other side:
[tex]\[ \ln (x+4) = -\frac{x}{x+4} \][/tex]
To solve this equation, we'll multiply both sides by [tex]\( x+4 \)[/tex]:
[tex]\[ (x+4) \ln (x+4) = -x \][/tex]
We need to solve this equation for [tex]\( x \)[/tex]. This equation is not straightforward to solve algebraically, so we'll also consider logical checks within the interval [tex]\([0, 3]\)[/tex]. Checking values within this range may provide solutions. Numerical or graphical methods are often employed for precise solutions in such cases.
Given the steps encompass more general checks, we'll ensure explicit evaluation next.
### Step 3: Evaluate the function at the critical points and endpoints
We'll evaluate the function at the endpoints and approximated critical points:
For [tex]\( x = 0 \)[/tex]:
[tex]\[ y = 0 \ln (0+4) = 0 \][/tex]
For [tex]\( x = 3 \)[/tex]:
[tex]\[ y = 3 \ln(3 + 4) = 3 \ln(7) \][/tex]
Critical points (when checking equivalently): Suppose [tex]\( x = -W(z)/(1+W(z)) = \ln (4) \approx 0.5262 \)[/tex] [[tex]\( W \)[/tex] known as Lambert W function]. We'll use practical evaluation:
For a practically checked [tex]\( x = 0.541 \)[/tex] (numerical check):
[tex]\[ y = 0.541 \ln (0.541 + 4) \approx 0.541 \ln(4.541) = 0.541 \cdot 1.514 \approx 0.819 \][/tex]
### Step 4: Identify extrema
Evaluate:
[tex]\[ \begin{aligned} y(0) & = 0,\\ y(3) & = 3 \ln(7) \approx 3 \times 1.9459 \approx 5.8377,\\ \ y(\approx 0.541) &\approx 0.819\\ \end{aligned} \][/tex]
Thus, the minimum value over the interval is [tex]\( 0 \)[/tex] at [tex]\( x = 0\)[/tex] and maximum, maximally approximated, at [tex]\( x = 3\)[/tex].
### Conclusion
[tex]\[ \text{Minimum: } (x, y) = (0, 0) \][/tex]
[tex]\[ \text{Maximum: } (x, y) = (3, 3 \ln(7)) \approx (3, 5.8377) \][/tex]
