Problem 32
Question
Use any method to find the relative extrema of the function \(f\). $$f(x)=x^{2} e^{x}$$
Step-by-Step Solution
Verified Answer
The function has a relative minimum at \(x = 0\) and a relative maximum at \(x = -2\).
1Step 1: Find the First Derivative
To find the relative extrema, we must first find the critical points. Start by finding the first derivative of the function using the product rule. The function is \(f(x) = x^2 e^x\). The derivative of \(x^2\) is \(2x\), and the derivative of \(e^x\) is \(e^x\).Thus, applying the product rule, we get:\[f'(x) = (x^2)' e^x + x^2 (e^x)' = 2x e^x + x^2 e^x = e^x(2x + x^2)\]
2Step 2: Set the Derivative Equal to Zero
To find the critical points, set the first derivative equal to zero and solve for \(x\):\[ e^x(2x + x^2) = 0 \]Since \(e^x\) is never zero, we focus on solving:\[2x + x^2 = 0\]
3Step 3: Solve the Quadratic Equation
Factor the quadratic equation:\[ x(x + 2) = 0 \]Thus, \(x = 0\) or \(x = -2\) are the solutions.
4Step 4: Test Critical Points
To determine whether these critical points are relative maxima or minima, use the second derivative test. First, find the second derivative of \(f(x)\):The first derivative was \(f'(x) = e^x(x^2 + 2x)\). Differentiate again:Use the product rule for \(e^x(x^2 + 2x)\).Let \(u = x^2 + 2x\) and \(v = e^x\).\(u' = 2x + 2\) and \(v' = e^x\).Applying the product rule:\[ f''(x) = (x^2 + 2x)e^x + e^x(2x + 2) = e^x(x^2 + 2x + 2x + 2) \]\[ f''(x) = e^x(x^2 + 4x + 2) \]Evaluate \(f''(x)\) at \(x = 0\) and \(x = -2\).At \(x = 0\):\[ f''(0) = e^0(0^2 + 4(0) + 2) = 2 \]The second derivative is positive at \(x = 0\), indicating a relative minimum.At \(x = -2\):\[ f''(-2) = e^{-2}((-2)^2 + 4(-2) + 2) = e^{-2}(4 - 8 + 2) = -2e^{-2} \]The second derivative is negative at \(x = -2\), indicating a relative maximum.
Key Concepts
Relative ExtremaProduct RuleSecond Derivative TestQuadratic Equation
Relative Extrema
Relative extrema refer to the high and low points of a function within a specific interval. These points can be either relative maxima or relative minima. Finding these points involves identifying where the first derivative changes sign.
- **Relative Maxima**: This is where the function changes from increasing to decreasing, indicating a peak.
- **Relative Minima**: This is where the function changes from decreasing to increasing, indicating a trough.
When solving for relative extrema, it is crucial first to calculate the function's critical points. Critical points occur where the first derivative equals zero or is undefined. Once the critical points are identified, tests like the second derivative test can determine whether these points are maxima, minima, or neither.
- **Relative Maxima**: This is where the function changes from increasing to decreasing, indicating a peak.
- **Relative Minima**: This is where the function changes from decreasing to increasing, indicating a trough.
When solving for relative extrema, it is crucial first to calculate the function's critical points. Critical points occur where the first derivative equals zero or is undefined. Once the critical points are identified, tests like the second derivative test can determine whether these points are maxima, minima, or neither.
Product Rule
The product rule is a fundamental calculus technique used to find the derivative of a product of two functions. If you have two functions, say \( u(x) \) and \( v(x) \), both of which are differentiable, the product rule states:
\[ (uv)' = u'v + uv'\]
In context, if you're differentiating a function like \( f(x) = x^2e^x \), apply the product rule as follows:
- Differentiate \( x^2 \) to get \( 2x \), and keep \( e^x \).
- Differentiate \( e^x \) to get \( e^x \), and keep \( x^2 \).
So the derivative using the product rule is:
\[ f'(x) = (x^2)'e^x + x^2(e^x)' = 2xe^x + x^2e^x = e^x(2x + x^2) \]
This method is essential for differentiating complex expressions involving the product of different functions.
\[ (uv)' = u'v + uv'\]
In context, if you're differentiating a function like \( f(x) = x^2e^x \), apply the product rule as follows:
- Differentiate \( x^2 \) to get \( 2x \), and keep \( e^x \).
- Differentiate \( e^x \) to get \( e^x \), and keep \( x^2 \).
So the derivative using the product rule is:
\[ f'(x) = (x^2)'e^x + x^2(e^x)' = 2xe^x + x^2e^x = e^x(2x + x^2) \]
This method is essential for differentiating complex expressions involving the product of different functions.
Second Derivative Test
The second derivative test is a useful method to determine the nature of the critical points found through the first derivative. Once you have a critical point, use the second derivative to assess whether it represents a relative maxima or minima.
The process involves:
- **Calculating the Second Derivative**: Differentiate the first derivative to obtain the second derivative \( f''(x) \).
- **Evaluating at Critical Points**: Plug the critical points into the second derivative.
- **Interpreting Results**:
- If \( f''(x) > 0 \), the function is concave up, indicating a relative minimum.
- If \( f''(x) < 0 \), the function is concave down, indicating a relative maximum.
In the original exercise, the second derivative test revealed that at \( x = 0 \), there's a relative minimum, and at \( x = -2 \), there's a relative maximum. This information provides essential insights into the function's behavior around these points.
The process involves:
- **Calculating the Second Derivative**: Differentiate the first derivative to obtain the second derivative \( f''(x) \).
- **Evaluating at Critical Points**: Plug the critical points into the second derivative.
- **Interpreting Results**:
- If \( f''(x) > 0 \), the function is concave up, indicating a relative minimum.
- If \( f''(x) < 0 \), the function is concave down, indicating a relative maximum.
In the original exercise, the second derivative test revealed that at \( x = 0 \), there's a relative minimum, and at \( x = -2 \), there's a relative maximum. This information provides essential insights into the function's behavior around these points.
Quadratic Equation
A quadratic equation is a second-degree polynomial equation in the form \( ax^2 + bx + c = 0 \). Solving quadratic equations is a fundamental algebraic skill required in calculus and other advanced mathematics.
For example, in the original exercise, the critical points were found by solving the quadratic equation:
\[ 2x + x^2 = 0 \]
This equation was factored as:
\[ x(x + 2) = 0 \]
Solving gives the roots \( x = 0 \) and \( x = -2 \). These roots correspond to the critical points of the function.
Quadratic equations can often be solved by:
For example, in the original exercise, the critical points were found by solving the quadratic equation:
\[ 2x + x^2 = 0 \]
This equation was factored as:
\[ x(x + 2) = 0 \]
Solving gives the roots \( x = 0 \) and \( x = -2 \). These roots correspond to the critical points of the function.
Quadratic equations can often be solved by:
- Factoring into a product of linear terms, as seen in this exercise.
- Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), when factoring is challenging.
Other exercises in this chapter
Problem 31
Use any method to find the relative extrema of the function \(f\). $$f(x)=\ln \left(1+x^{2}\right)$$
View solution Problem 32
In each part sketch a continuous curve \(y=f(x)\) with the stated properties. (a) \(f(2)=4, \quad f^{\prime}(2)=0, \quad f^{\prime \prime}(x)0\) for \(x2\) (c)
View solution Problem 33
In each part, assume that \(a\) is a constant and find the inflection points, if any. (a) \(f(x)=(x-a)^{3}\) (b) \(f(x)=(x-a)^{4}\)
View solution Problem 33
Give a complete graph of the function, and identify the location of all relative extrema and inflection points. Check your work with a graphing utility. $$x+\si
View solution