Problem 61
Question
Graph the polynomial and determine how many local maxima and minima it has. $$y=x^{3}-x^{2}-x$$
Step-by-Step Solution
Verified Answer
There is one local maximum and one local minimum.
1Step 1: Find the derivative
To identify local maxima and minima, we need to find the critical points of the function by taking the derivative. The derivative of the polynomial \(y = x^3 - x^2 - x\) is obtained as follows:\[\frac{dy}{dx} = 3x^2 - 2x - 1\]
2Step 2: Set the derivative to zero
Critical points occur where the derivative is zero. Set \(3x^2 - 2x - 1 = 0\) and solve for \(x\). This is a quadratic equation that can be solved using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
3Step 3: Solve the quadratic equation
Plug in the values \(a = 3\), \(b = -2\), and \(c = -1\) into the quadratic formula:\[x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3}\]\[x = \frac{2 \pm \sqrt{4 + 12}}{6}\]\[x = \frac{2 \pm \sqrt{16}}{6}\]\[x = \frac{2 \pm 4}{6}\]Thus, \(x = 1\) and \(x = -\frac{1}{3}\).
4Step 4: Test intervals between critical points
Determine whether each critical point is a local maximum or minimum by testing the sign of the derivative in the intervals around the critical points. Test points like \(x = -1\), \(x = 0\), and \(x = 2\).
5Step 5: Use the first derivative test
For \(x = -1\), plug into the derivative:\[3(-1)^2 - 2(-1) - 1 = 3 + 2 - 1 = 4\] (positive).For \(x = 0\), plug into the derivative:\[3(0)^2 - 2(0) - 1 = -1\] (negative).For \(x = 2\), plug into the derivative:\[3(2)^2 - 2(2) - 1 = 12 - 4 - 1 = 7\] (positive).Since the derivative changes from positive to negative at \(x = -\frac{1}{3}\), there is a local maximum. Since the derivative changes from negative to positive at \(x = 1\), there is a local minimum.
6Step 6: Graph the polynomial for confirmation
Sketch the graph of \(y = x^3 - x^2 - x\) using the critical points and derivative sign to confirm these local extrema interpretations. The graph confirms a local maximum at \(x = -\frac{1}{3}\) and a local minimum at \(x = 1\).
Key Concepts
Critical PointsDerivativeLocal ExtremaQuadratic Formula
Critical Points
Critical points of a polynomial function play a crucial role in understanding the behavior of its graph.
They are the points where the derivative of the function is either zero or undefined.
By identifying these points, we can determine where the function may have local maxima or minima.
For a polynomial function like \(y = x^3 - x^2 - x\), the critical points occur where the derivative, \(3x^2 - 2x - 1\), equals zero.
Solving the equation \(3x^2 - 2x - 1 = 0\), results in potential critical points.
By knowing the values where the derivative equals zero, we can proceed to analyze the nature of these points using further tests like the First Derivative Test.
In our example, the critical points are determined as \(x = 1\) and \(x = -\frac{1}{3}\).
They are the points where the derivative of the function is either zero or undefined.
By identifying these points, we can determine where the function may have local maxima or minima.
For a polynomial function like \(y = x^3 - x^2 - x\), the critical points occur where the derivative, \(3x^2 - 2x - 1\), equals zero.
Solving the equation \(3x^2 - 2x - 1 = 0\), results in potential critical points.
By knowing the values where the derivative equals zero, we can proceed to analyze the nature of these points using further tests like the First Derivative Test.
In our example, the critical points are determined as \(x = 1\) and \(x = -\frac{1}{3}\).
Derivative
The derivative of a function measures the rate at which the function's value changes relative to changes in the input.
In simpler terms, it provides the slope of the tangent line to the curve of the function at any given point.
For polynomial functions, finding the derivative is a straightforward but vital process.
Taking the derivative of a polynomial like \(y = x^3 - x^2 - x\) gives you \(\frac{dy}{dx} = 3x^2 - 2x - 1\).
This expression represents the slope of the function \(y\) at different values of \(x\).
By analyzing this derivative, you can identify the behavior of the polynomial such as rising, falling, or attaining extreme values at particular points.
A positive derivative suggests a rising slope, whereas a negative derivative suggests a falling slope.
In simpler terms, it provides the slope of the tangent line to the curve of the function at any given point.
For polynomial functions, finding the derivative is a straightforward but vital process.
Taking the derivative of a polynomial like \(y = x^3 - x^2 - x\) gives you \(\frac{dy}{dx} = 3x^2 - 2x - 1\).
This expression represents the slope of the function \(y\) at different values of \(x\).
By analyzing this derivative, you can identify the behavior of the polynomial such as rising, falling, or attaining extreme values at particular points.
A positive derivative suggests a rising slope, whereas a negative derivative suggests a falling slope.
Local Extrema
Local extrema refer to the points on a graph of a function where the function reaches a local maximum or a local minimum.
These points not only provide crucial insights into the nature of the function but also help in sketching its graph more precisely.
To determine the local extrema, we rely on analyzing the critical points, which can potentially be a local maximum, a local minimum, or neither.
The First Derivative Test is commonly used to study these points.
By observing how the sign of the derivative changes around the critical points, we can identify if they are maxima or minima.
For instance, at \(x = -\frac{1}{3}\) and \(x = 1\), checking intervals around these points showed a change from positive to negative and from negative to positive, indicating a local maximum and minimum, respectively.
Understanding local extrema allows for a deeper grasp of the function's overall shape and behavior.
These points not only provide crucial insights into the nature of the function but also help in sketching its graph more precisely.
To determine the local extrema, we rely on analyzing the critical points, which can potentially be a local maximum, a local minimum, or neither.
The First Derivative Test is commonly used to study these points.
By observing how the sign of the derivative changes around the critical points, we can identify if they are maxima or minima.
For instance, at \(x = -\frac{1}{3}\) and \(x = 1\), checking intervals around these points showed a change from positive to negative and from negative to positive, indicating a local maximum and minimum, respectively.
Understanding local extrema allows for a deeper grasp of the function's overall shape and behavior.
Quadratic Formula
The quadratic formula is a reliable method for finding the roots of quadratic equations. These roots are the \(x\)-values where the polynomial is equal to zero.
This formula is especially handy when factoring is difficult or not obvious.
For a standard quadratic equation expressed as \(ax^2 + bx + c = 0\), the quadratic formula is given by\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Using this formula entails substituting values of \(a\), \(b\), and \(c\), which represent coefficients from the quadratic equation.
In the exercise, solving \(3x^2 - 2x - 1 = 0\) through the quadratic formula helps pinpoint \(x\) values where the derivative is zero which are \(x = 1\) and \(x = -\frac{1}{3}\).
Using the quadratic formula simplifies the process of finding critical points for further analysis of the function's nature and behavior.
This formula is especially handy when factoring is difficult or not obvious.
For a standard quadratic equation expressed as \(ax^2 + bx + c = 0\), the quadratic formula is given by\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Using this formula entails substituting values of \(a\), \(b\), and \(c\), which represent coefficients from the quadratic equation.
In the exercise, solving \(3x^2 - 2x - 1 = 0\) through the quadratic formula helps pinpoint \(x\) values where the derivative is zero which are \(x = 1\) and \(x = -\frac{1}{3}\).
Using the quadratic formula simplifies the process of finding critical points for further analysis of the function's nature and behavior.
Other exercises in this chapter
Problem 61
Find all the real zeros of the polynomial. Use the quadratic formula if necessary, as in Example \(3(a)\) $$P(x)=x^{4}-5 x^{3}+6 x^{2}+4 x-8$$
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Find the local maximum and minimum values of the function and the value of \(x\) at which each occurs. State each answer correct to two decimal places. $$v(x)=\
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Find a polynomial of the specified degree that has the given zeros. Degree \(4 ; \quad\) zeros -1,1,3,5
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Find all solutions of the equation and express them in the form \(a+b i\) $$x^{2}-6 x+10=0$$
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