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To determine the maximum number of relative extrema for the function [tex]\( f(x) = 3x^4 - x^2 + 4x - 2 \)[/tex], we can follow these detailed steps:
### Step 1: Find the first derivative of the function
The first derivative, [tex]\( f'(x) \)[/tex], helps us identify critical points where the function's slope is zero or undefined.
Given:
[tex]\[ f(x) = 3x^4 - x^2 + 4x - 2 \][/tex]
The first derivative is:
[tex]\[ f'(x) = \frac{d}{dx}(3x^4 - x^2 + 4x - 2) = 12x^3 - 2x + 4 \][/tex]
### Step 2: Find the critical points
To find the critical points, we set the first derivative equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ 12x^3 - 2x + 4 = 0 \][/tex]
The solutions to this equation are the critical points. Solving this cubic equation, we obtain the critical points:
[tex]\[ x = \left[ -\frac{1}{6(-\frac{1}{2} - \frac{\sqrt{3}i}{2}) \left(\frac{\sqrt{322}}{4} + \frac{9}{2}\right)^{1/3}} - \left(-\frac{1}{2} - \frac{\sqrt{3}i}{2}\right) \left(\frac{\sqrt{322}}{4} + \frac{9}{2}\right)^{1/3}/3, -\left(-\frac{1}{2} + \frac{\sqrt{3}i}{2}\right) \left(\frac{\sqrt{322}}{4} + \frac{9}{2}\right)^{1/3}/3 - \frac{1}{6(-\frac{1}{2} + \frac{\sqrt{3}i}{2})(\frac{\sqrt{322}}{4} + \frac{9}{2})^{1/3}}, -(\frac{\sqrt{322}}{4} + \frac{9}{2})^{1/3}/3 - \frac{1}{6(\frac{\sqrt{322}}{4} + \frac{9}{2})^{1/3}} \right] \][/tex]
### Step 3: Determine the nature of the critical points using the second derivative
The second derivative, [tex]\( f''(x) \)[/tex], tells us about the concavity of the function at the critical points, helping us categorize them as relative minima, maxima, or points of inflection.
The second derivative is:
[tex]\[ f''(x) = \frac{d}{dx}(12x^3 - 2x + 4) = 36x^2 - 2 \][/tex]
We substitute each critical point into the second derivative to determine its sign:
[tex]\[ f''(-\frac{1}{6(-\frac{1}{2} - \frac{\sqrt{3}i}{2})(\frac{\sqrt{322}}{4} + \frac{9}{2})^{1/3}} - (-\frac{1}{2} - \frac{\sqrt{3}i}{2})(\frac{\sqrt{322}}{4} + \frac{9}{2})^{1/3}/3) \ne 0 \][/tex]
[tex]\[ f''(-(-\frac{1}{2} + \frac{\sqrt{3}i}{2})(\frac{\sqrt{322}}{4} + \frac{9}{2})^{1/3}/3 - \frac{1}{6(-\frac{1}{2} + \frac{\sqrt{3}i}{2}) (\sqrt{322}/4 + \frac{9}{2})^{1/3})} \ne 0 \][/tex]
[tex]\[ f''(-(\frac{\sqrt{322}}{4} + \frac{9}{2})^{1/3}/3 - \frac{1}{6(\frac{\sqrt{322}}{4} + \frac{9}{2})^{1/3}} \ne 0 \][/tex]
Since the second derivative at each critical point is non-zero, each of these points is a point of relative extrema (either a relative maximum or minimum).
### Step 4: Count the relative extrema
The number of valid critical points for which the second derivative is non-zero represents the maximum number of relative extrema. In this case, we have identified three such critical points.
### Conclusion
Thus, the maximum number of relative extrema contained in the graph of the function [tex]\( f(x) = 3x^4 - x^2 + 4x - 2 \)[/tex] is:
[tex]\[ \boxed{3} \][/tex]
### Step 1: Find the first derivative of the function
The first derivative, [tex]\( f'(x) \)[/tex], helps us identify critical points where the function's slope is zero or undefined.
Given:
[tex]\[ f(x) = 3x^4 - x^2 + 4x - 2 \][/tex]
The first derivative is:
[tex]\[ f'(x) = \frac{d}{dx}(3x^4 - x^2 + 4x - 2) = 12x^3 - 2x + 4 \][/tex]
### Step 2: Find the critical points
To find the critical points, we set the first derivative equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ 12x^3 - 2x + 4 = 0 \][/tex]
The solutions to this equation are the critical points. Solving this cubic equation, we obtain the critical points:
[tex]\[ x = \left[ -\frac{1}{6(-\frac{1}{2} - \frac{\sqrt{3}i}{2}) \left(\frac{\sqrt{322}}{4} + \frac{9}{2}\right)^{1/3}} - \left(-\frac{1}{2} - \frac{\sqrt{3}i}{2}\right) \left(\frac{\sqrt{322}}{4} + \frac{9}{2}\right)^{1/3}/3, -\left(-\frac{1}{2} + \frac{\sqrt{3}i}{2}\right) \left(\frac{\sqrt{322}}{4} + \frac{9}{2}\right)^{1/3}/3 - \frac{1}{6(-\frac{1}{2} + \frac{\sqrt{3}i}{2})(\frac{\sqrt{322}}{4} + \frac{9}{2})^{1/3}}, -(\frac{\sqrt{322}}{4} + \frac{9}{2})^{1/3}/3 - \frac{1}{6(\frac{\sqrt{322}}{4} + \frac{9}{2})^{1/3}} \right] \][/tex]
### Step 3: Determine the nature of the critical points using the second derivative
The second derivative, [tex]\( f''(x) \)[/tex], tells us about the concavity of the function at the critical points, helping us categorize them as relative minima, maxima, or points of inflection.
The second derivative is:
[tex]\[ f''(x) = \frac{d}{dx}(12x^3 - 2x + 4) = 36x^2 - 2 \][/tex]
We substitute each critical point into the second derivative to determine its sign:
[tex]\[ f''(-\frac{1}{6(-\frac{1}{2} - \frac{\sqrt{3}i}{2})(\frac{\sqrt{322}}{4} + \frac{9}{2})^{1/3}} - (-\frac{1}{2} - \frac{\sqrt{3}i}{2})(\frac{\sqrt{322}}{4} + \frac{9}{2})^{1/3}/3) \ne 0 \][/tex]
[tex]\[ f''(-(-\frac{1}{2} + \frac{\sqrt{3}i}{2})(\frac{\sqrt{322}}{4} + \frac{9}{2})^{1/3}/3 - \frac{1}{6(-\frac{1}{2} + \frac{\sqrt{3}i}{2}) (\sqrt{322}/4 + \frac{9}{2})^{1/3})} \ne 0 \][/tex]
[tex]\[ f''(-(\frac{\sqrt{322}}{4} + \frac{9}{2})^{1/3}/3 - \frac{1}{6(\frac{\sqrt{322}}{4} + \frac{9}{2})^{1/3}} \ne 0 \][/tex]
Since the second derivative at each critical point is non-zero, each of these points is a point of relative extrema (either a relative maximum or minimum).
### Step 4: Count the relative extrema
The number of valid critical points for which the second derivative is non-zero represents the maximum number of relative extrema. In this case, we have identified three such critical points.
### Conclusion
Thus, the maximum number of relative extrema contained in the graph of the function [tex]\( f(x) = 3x^4 - x^2 + 4x - 2 \)[/tex] is:
[tex]\[ \boxed{3} \][/tex]
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