Problem 39
Question
Use any method to find the relative extrema of the function \(f .\) $$ f(x)=x^{3}(x+1)^{2} $$
Step-by-Step Solution
Verified Answer
The relative extrema are at \(x = 0\), \(x = -1\), and \(x = -\frac{3}{5}\).
1Step 1: Identify the Function
The given function is \( f(x) = x^3(x+1)^2 \). We want to find the relative extrema of this function, which occur where the first derivative is zero or undefined.
2Step 2: Find the First Derivative
To find the critical points, first find the derivative of the function. Use the product rule:\[(fg)' = f'g + fg'\]Let \( f(x) = x^3 \) and \( g(x) = (x+1)^2 \), then:\[ f'(x) = 3x^2 \quad \text{and} \quad g'(x) = 2(x+1) \]Applying the product rule gives:\[(x^3)'(x+1)^2 + x^3((x+1)^2)' = 3x^2(x+1)^2 + x^3 \cdot 2(x+1)\]Simplifying, we get:\[ f'(x) = 3x^2(x+1)^2 + 2x^3(x+1) \]
3Step 3: Simplify the Derivative
Factor the derivative to find critical points easier:\[ f'(x) = x^2(x+1)(3(x+1) + 2x) \]Further simplification leads to:\[ f'(x) = x^2(x+1)(3x + 3 + 2x) = x^2(x+1)(5x + 3)\]
4Step 4: Find Critical Points
Set the derivative equal to zero to find critical points:\[ f'(x) = x^2(x+1)(5x + 3) = 0 \]This gives the solutions:- \( x^2 = 0 \), hence \( x = 0 \),- \( x+1 = 0 \), hence \( x = -1 \),- \( 5x + 3 = 0 \), solving gives:\[ x = -\frac{3}{5} \].
5Step 5: Test Critical Points for Extrema
Use the second derivative test or test values around the critical points in the first derivative to determine if these points are relative minima, maxima, or neither. - Test around \(x = 0\), \(x = -1\), and \(x = -\frac{3}{5}\). - At \(x = 0\), test with values \(x = -0.1\) and \(x = 0.1\). - At \(x = -1\), use \(x = -1.1\) and \(x = -0.9\). - At \(x = -\frac{3}{5}\), choose \(x = -0.6\) and \(x = -0.7\).Determine the behavior around each critical point to classify it.
Key Concepts
Relative ExtremaDerivativeCritical PointsProduct Rule
Relative Extrema
Relative extrema refer to the peaks and valleys of a function on a graph. They are the points where a function
changes from increasing to decreasing or vice versa. These points are of particular interest as they can represent maximum
and minimum values in a certain interval. The importance of identifying relative extrema lies in their wide application across
different fields, such as physics, economics, and engineering.
To find the relative extrema of a function, we look for critical points, which occur at locations where the slope is zero, or the derivative of the function doesn't exist. Once the critical points are determined, they can be tested to ascertain if they represent relative minima or maxima using methods like the first derivative test or the second derivative test.
To find the relative extrema of a function, we look for critical points, which occur at locations where the slope is zero, or the derivative of the function doesn't exist. Once the critical points are determined, they can be tested to ascertain if they represent relative minima or maxima using methods like the first derivative test or the second derivative test.
Derivative
The derivative of a function is a fundamental concept in calculus that represents the rate of change of a function with respect to a variable. If we imagine a graph of a function, the derivative at a given point can be visualized as the slope of the tangent line at that point.
For the function in our exercise, the derivative helps us understand how the function behaves in terms of increasing and decreasing values. Calculating the derivative is the first step in identifying the critical points and subsequently the relative extrema of the function. The notation for the derivative of a function \( f(x) \) is \( f'(x) \). Computing this can involve applying different rules, such as the product rule, especially in functions that are products of multiple terms.
For the function in our exercise, the derivative helps us understand how the function behaves in terms of increasing and decreasing values. Calculating the derivative is the first step in identifying the critical points and subsequently the relative extrema of the function. The notation for the derivative of a function \( f(x) \) is \( f'(x) \). Computing this can involve applying different rules, such as the product rule, especially in functions that are products of multiple terms.
Critical Points
Critical points of a function occur where the derivative is zero or undefined. These points signify the locations where the function might have a relative minimum, relative maximum, or a plateau (level point without a local peak or valley). Identifying these points is crucial for sketching a graph and determining important behaviors of the function.
In the provided exercise, the critical points are found by setting the derivative equal to zero and solving for the value of \( x \). For example, after simplifying the derivative, \( f'(x) = x^2(x+1)(5x + 3) = 0 \), we derive the points \( x = 0 \), \( x = -1 \), and \( x = -\frac{3}{5} \). Once these values are obtained, further testing, like the second derivative test, may be necessary to confirm whether these are points of relative extrema.
In the provided exercise, the critical points are found by setting the derivative equal to zero and solving for the value of \( x \). For example, after simplifying the derivative, \( f'(x) = x^2(x+1)(5x + 3) = 0 \), we derive the points \( x = 0 \), \( x = -1 \), and \( x = -\frac{3}{5} \). Once these values are obtained, further testing, like the second derivative test, may be necessary to confirm whether these are points of relative extrema.
Product Rule
When a function is expressed as the product of two other functions, the product rule is a technique used to find the derivative of the original function. The rule is stated as: if \( f(x) = u(x)v(x) \), then \( f'(x) = u'(x)v(x) + u(x)v'(x) \).
In our example, the function \( f(x) = x^3(x+1)^2 \) is made of two parts, \( u(x) = x^3 \) and \( v(x) = (x+1)^2 \). The product rule allows us to find the derivative \( f'(x) \) by differentiating each function separately and then applying the rule. This gives us insight into the behavior of the function as a whole, which is crucial for identifying critical points and understanding where the function may achieve relative extrema.
In our example, the function \( f(x) = x^3(x+1)^2 \) is made of two parts, \( u(x) = x^3 \) and \( v(x) = (x+1)^2 \). The product rule allows us to find the derivative \( f'(x) \) by differentiating each function separately and then applying the rule. This gives us insight into the behavior of the function as a whole, which is crucial for identifying critical points and understanding where the function may achieve relative extrema.
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