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Magazine Chapter 3: Problem 13
Find the relative extrema and points o[ inflection of the [unction. List your answers in terms of ordered pairs. Use the Second-Derivative Test, where possible. Then sketch the graph. $$ f(x)=3 x^{4}+4 x^{3} $$
Short Answer
Expert verified
Relative minimum at (-1, -1). Inflection points at (0, 0) and (-2/3, -16/27).
Step by step solution
01
Find the First Derivative
To find the relative extrema, begin by finding the first derivative of the function. For the function \find f(x) = 3x^4 + 4x^3, apply the power rule to each term:\(f'(x) = 12x^3 + 12x^2\)
02
Set the First Derivative to Zero
To find critical points, set the first derivative equal to zero and solve for x:\(12x^3 + 12x^2 = 0\)Factor out the common term:\(12x^2(x + 1) = 0\)So, the critical points are:\(x = 0 \text{ and } x = -1\)
03
Find the Second Derivative
To determine the nature of the critical points, find the second derivative of the function:\(f''(x) = 36x^2 + 24x\)
04
Apply the Second Derivative Test
Evaluate the second derivative at each critical point:For \(x = -1\),\( f''(-1) = 36(-1)^2 + 24(-1) = 36 - 24 = 12\).Since \(f''(-1) > 0\), the function has a relative minimum at \(x = -1\) For \(x = 0\),\( f''(0) = 36(0)^2 + 24(0) = 0\), the second derivative test is inconclusive here.
05
Determine Inflection Points
To find points of inflection, set the second derivative equal to zero and solve for x:\(36x^2 + 24x = 0\)Factor out the common term:\(12x(3x + 2) = 0\)So, the points are:\(x = 0 \text{ and } x = -\frac{2}{3}\)
06
Evaluate f(x) at Critical and Inflection Points
Find the function values at the critical and inflection points:\(f(0) = 3(0)^4 + 4(0)^3 = 0\)\(f(-1) = 3(-1)^4 + 4(-1)^3 = 3 - 4 = -1\) \(f(-\frac{2}{3}) = 3(-\frac{2}{3})^4 + 4(-\frac{2}{3})^3\)\(= 3\left(\frac{16}{81}\right) + 4\left(\frac{-8}{27}\right)\)\(= \frac{48}{81} - \frac{32}{27}\)\(= \frac{48}{81} - \frac{96}{81}\)\(= \frac{-48}{81}\)\(= -\frac{16}{27}\)
07
List the Extremes and Inflection Points
For the relative extrema and inflection points, we have:Relative minimum at \((-1, -1)\)Inflection points at \((0, 0)\) and \((-\frac{2}{3}, -\frac{16}{27})\)
08
Sketch the Graph
Plot the function using the critical and inflection points to understand the shape of the function. The function has a minimum at \((-1, -1)\) and changes concavity at \((0, 0)\) and \((-\frac{2}{3}, -\frac{16}{27})\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
relative extrema
Relative extrema are the points where a function reaches its local maximum or minimum values. In simpler terms, these are the peaks and valleys of the function within a given interval. To find them:
In our problem, we found the first derivative: \( f'(x) = 12x^3 + 12x^2 \). By setting it to zero, we got critical points: \( x = 0 \) and \( x = -1 \). For \( x = -1 \), the second derivative \( f''(-1) \) was positive, confirming a relative minimum at \( (-1, -1) \). The second derivative test was inconclusive for \( x = 0 \).
- Calculate the first derivative of the function.
- Set the first derivative equal to zero to find the critical points.
- Use the second derivative test to determine whether each critical point is a relative maximum or minimum.
In our problem, we found the first derivative: \( f'(x) = 12x^3 + 12x^2 \). By setting it to zero, we got critical points: \( x = 0 \) and \( x = -1 \). For \( x = -1 \), the second derivative \( f''(-1) \) was positive, confirming a relative minimum at \( (-1, -1) \). The second derivative test was inconclusive for \( x = 0 \).
points of inflection
Points of inflection are points where the function changes concavity from concave up to concave down or vice versa. To find them:
In our example, the second derivative is \( f''(x) = 36x^2 + 24x \). Solving \( 36x^2 + 24x = 0 \) gave us potential inflection points at \( x = 0 \) and \( x = -\frac{2}{3} \). Evaluations confirmed that concavity changes at these points: \( (0,0) \) and \( (-\frac{2}{3}, -\frac{16}{27}) \).
- Calculate the second derivative.
- Set the second derivative equal to zero to find potential points of inflection.
- Verify if the concavity changes at these points by evaluating the second derivative around them.
In our example, the second derivative is \( f''(x) = 36x^2 + 24x \). Solving \( 36x^2 + 24x = 0 \) gave us potential inflection points at \( x = 0 \) and \( x = -\frac{2}{3} \). Evaluations confirmed that concavity changes at these points: \( (0,0) \) and \( (-\frac{2}{3}, -\frac{16}{27}) \).
first derivative
The first derivative of a function, denoted \( f'(x) \), is a crucial tool in calculus for determining the slope of the tangent line at any point on the function. It helps in identifying where the function is increasing or decreasing, and locating relative extrema.
For the problem at hand, our function is \( f(x) = 3x^4 + 4x^3 \). Applying the power rule gave us \( f'(x) = 12x^3 + 12x^2 \). Setting this equal to zero, \( 12x^3 + 12x^2 = 0 \), and factoring out common terms, \( 12x^2(x + 1) = 0 \), we found critical points \( x = 0 \) and \( x = -1 \).
For the problem at hand, our function is \( f(x) = 3x^4 + 4x^3 \). Applying the power rule gave us \( f'(x) = 12x^3 + 12x^2 \). Setting this equal to zero, \( 12x^3 + 12x^2 = 0 \), and factoring out common terms, \( 12x^2(x + 1) = 0 \), we found critical points \( x = 0 \) and \( x = -1 \).
second derivative
The second derivative, denoted \( f''(x) \), examines the concavity of the function. It helps us determine whether critical points found from the first derivative are maxima, minima, or points of inflection.
In our example, the first derivative is \( f'(x) = 12x^3 + 12x^2 \). Differentiating again, we get the second derivative: \( f''(x) = 36x^2 + 24x \). We use this to apply the second derivative test at the critical points. For \( x = -1 \), \( f''(-1) = 12 \), which is positive, indicating a relative minimum at \( (-1, -1) \). For \( x = 0 \), \( f''(0) = 0 \), leaving the test inconclusive.
In our example, the first derivative is \( f'(x) = 12x^3 + 12x^2 \). Differentiating again, we get the second derivative: \( f''(x) = 36x^2 + 24x \). We use this to apply the second derivative test at the critical points. For \( x = -1 \), \( f''(-1) = 12 \), which is positive, indicating a relative minimum at \( (-1, -1) \). For \( x = 0 \), \( f''(0) = 0 \), leaving the test inconclusive.
critical points
Critical points are where the first derivative of a function is zero or undefined. These points are significant because they potentially represent local maxima, minima, or points of inflection.
To find critical points:
In our problem, we derived \( f'(x) = 12x^3 + 12x^2 \). Setting it to zero gave us \( x = 0 \) and \( x = -1 \) as critical points. Specifically, \( x = -1 \) was confirmed as a relative minimum using the second derivative test, while \( x = 0 \) needed further analysis due to the test being inconclusive.
To find critical points:
- Compute the first derivative of the function.
- Set this derivative equal to zero.
- Solve for the values of \( x \).
In our problem, we derived \( f'(x) = 12x^3 + 12x^2 \). Setting it to zero gave us \( x = 0 \) and \( x = -1 \) as critical points. Specifically, \( x = -1 \) was confirmed as a relative minimum using the second derivative test, while \( x = 0 \) needed further analysis due to the test being inconclusive.
