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Sagot :
To find the maximum number of relative extrema of the function [tex]\( f(x) = 3x^5 - x^3 + 4x - 2 \)[/tex], we need to investigate where the first derivative changes sign. The process involves finding the first derivative, solving for critical points, and determining the nature of these points. Let's go through this step-by-step.
Step 1: Find the first derivative of the function.
Given the function:
[tex]\[ f(x) = 3x^5 - x^3 + 4x - 2 \][/tex]
We calculate the first derivative:
[tex]\[ f'(x) = \frac{d}{dx}(3x^5) - \frac{d}{dx}(x^3) + \frac{d}{dx}(4x) - \frac{d}{dx}(2) \][/tex]
[tex]\[ f'(x) = 15x^4 - 3x^2 + 4 \][/tex]
Step 2: Solve for the critical points by setting the first derivative to zero.
Set [tex]\( f'(x) = 0 \)[/tex]:
[tex]\[ 15x^4 - 3x^2 + 4 = 0 \][/tex]
Step 3: Solve the equation for [tex]\( x \)[/tex].
This is a polynomial equation of degree 4. To solve it, let [tex]\( y = x^2 \)[/tex]. Then the equation becomes:
[tex]\[ 15y^2 - 3y + 4 = 0 \][/tex]
This equation can be solved using the quadratic formula [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 15 \)[/tex], [tex]\( b = -3 \)[/tex], and [tex]\( c = 4 \)[/tex].
[tex]\[ y = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 15 \cdot 4}}{2 \cdot 15} \][/tex]
[tex]\[ y = \frac{3 \pm \sqrt{9 - 240}}{30} \][/tex]
[tex]\[ y = \frac{3 \pm \sqrt{-231}}{30} \][/tex]
Since the discriminant [tex]\( -231 \)[/tex] is negative, there are no real solutions for [tex]\( y \)[/tex]. This implies that there are no real critical points for [tex]\( x \)[/tex], because the quadratic equation does not cross the x-axis.
Conclusion:
Since there are no real critical points where the first derivative equals zero, there are no points where the derivative changes sign. Thus, the function [tex]\( f(x) = 3x^5 - x^3 + 4x - 2 \)[/tex] does not have any relative extrema (neither maxima nor minima) on the real number line.
Therefore, the maximum number of relative extrema contained in the graph of this function is:
[tex]\[ \boxed{0} \][/tex]
Step 1: Find the first derivative of the function.
Given the function:
[tex]\[ f(x) = 3x^5 - x^3 + 4x - 2 \][/tex]
We calculate the first derivative:
[tex]\[ f'(x) = \frac{d}{dx}(3x^5) - \frac{d}{dx}(x^3) + \frac{d}{dx}(4x) - \frac{d}{dx}(2) \][/tex]
[tex]\[ f'(x) = 15x^4 - 3x^2 + 4 \][/tex]
Step 2: Solve for the critical points by setting the first derivative to zero.
Set [tex]\( f'(x) = 0 \)[/tex]:
[tex]\[ 15x^4 - 3x^2 + 4 = 0 \][/tex]
Step 3: Solve the equation for [tex]\( x \)[/tex].
This is a polynomial equation of degree 4. To solve it, let [tex]\( y = x^2 \)[/tex]. Then the equation becomes:
[tex]\[ 15y^2 - 3y + 4 = 0 \][/tex]
This equation can be solved using the quadratic formula [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 15 \)[/tex], [tex]\( b = -3 \)[/tex], and [tex]\( c = 4 \)[/tex].
[tex]\[ y = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 15 \cdot 4}}{2 \cdot 15} \][/tex]
[tex]\[ y = \frac{3 \pm \sqrt{9 - 240}}{30} \][/tex]
[tex]\[ y = \frac{3 \pm \sqrt{-231}}{30} \][/tex]
Since the discriminant [tex]\( -231 \)[/tex] is negative, there are no real solutions for [tex]\( y \)[/tex]. This implies that there are no real critical points for [tex]\( x \)[/tex], because the quadratic equation does not cross the x-axis.
Conclusion:
Since there are no real critical points where the first derivative equals zero, there are no points where the derivative changes sign. Thus, the function [tex]\( f(x) = 3x^5 - x^3 + 4x - 2 \)[/tex] does not have any relative extrema (neither maxima nor minima) on the real number line.
Therefore, the maximum number of relative extrema contained in the graph of this function is:
[tex]\[ \boxed{0} \][/tex]
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