Answer :
To determine whether the polynomial [tex]\( g(x) = 2x^4 - 5x^3 + 3x^2 - 2x + 1 \)[/tex] has any local minima or maxima, we follow these steps:
1. Find the first derivative [tex]\( g'(x) \)[/tex]:
This will help identify the critical points where the function's slope is zero.
[tex]\[ g'(x) = \frac{d}{dx}(2x^4 - 5x^3 + 3x^2 - 2x + 1) = 8x^3 - 15x^2 + 6x - 2 \][/tex]
2. Solve [tex]\( g'(x) = 0 \)[/tex] to find critical points:
We solve the equation [tex]\( 8x^3 - 15x^2 + 6x - 2 = 0 \)[/tex] to find the critical points. The roots of this equation are the critical points. The solutions to this polynomial are:
[tex]\[ x_1 = \frac{5}{8} + \left( -\frac{1}{2} - \frac{\sqrt{3}i}{2} \right) \left( \frac{3\sqrt{7}}{64} + \frac{69}{512} \right)^{1/3} + \frac{9}{64 \left( -\frac{1}{2} - \frac{\sqrt{3}i}{2} \right) \left( \frac{3\sqrt{7}}{64} + \frac{69}{512} \right)^{1/3}} \][/tex]
[tex]\[ x_2 = \frac{5}{8} + \frac{9}{64 \left( -\frac{1}{2} + \frac{\sqrt{3}i}{2} \right) \left( \frac{3\sqrt{7}}{64} + \frac{69}{512} \right)^{1/3}} + \left( -\frac{1}{2} + \frac{\sqrt{3}i}{2} \right) \left( \frac{3\sqrt{7}}{64} + \frac{69}{512} \right)^{1/3} \][/tex]
[tex]\[ x_3 = \frac{9}{64 \left( \frac{3\sqrt{7}}{64} + \frac{69}{512} \right)^{1/3}} + \frac{5}{8} + \left( \frac{3\sqrt{7}}{64} + \frac{69}{512} \right)^{1/3} \][/tex]
3. Determine whether each critical point corresponds to a local maxima, minima, or neither:
To classify the critical points, we need to evaluate the second derivative [tex]\( g''(x) \)[/tex] at these points.
[tex]\[ g''(x) = \frac{d}{dx}(8x^3 - 15x^2 + 6x - 2) = 24x^2 - 30x + 6 \][/tex]
4. Evaluate [tex]\( g''(x) \)[/tex] at each critical point:
- At [tex]\( x_1 \)[/tex], [tex]\( g''(x_1) = -2.08248441322538 + 7.42836007216895i \)[/tex]
- At [tex]\( x_2 \)[/tex], [tex]\( g''(x_2) = -2.08248441322538 - 7.42836007216895i \)[/tex]
- At [tex]\( x_3 \)[/tex], [tex]\( g''(x_3) = 14.2899688264508 \)[/tex]
From these evaluations:
- Since [tex]\( g''(x_3) = 14.2899688264508 > 0 \)[/tex], [tex]\( x_3 \)[/tex] is a local minimum because a positive second derivative indicates that the function is concave up at this point.
- [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex] have complex values for their second derivatives, implying that these critical points are complex and thus not relevant for local minima or maxima in the real domain.
Hence, the polynomial function [tex]\( g(x) = 2x^4 - 5x^3 + 3x^2 - 2x + 1 \)[/tex] has a local minimum, not maxima.
Therefore, the correct answer is:
a. It has a local minima at approximately [tex]\( 1.65 \)[/tex].
1. Find the first derivative [tex]\( g'(x) \)[/tex]:
This will help identify the critical points where the function's slope is zero.
[tex]\[ g'(x) = \frac{d}{dx}(2x^4 - 5x^3 + 3x^2 - 2x + 1) = 8x^3 - 15x^2 + 6x - 2 \][/tex]
2. Solve [tex]\( g'(x) = 0 \)[/tex] to find critical points:
We solve the equation [tex]\( 8x^3 - 15x^2 + 6x - 2 = 0 \)[/tex] to find the critical points. The roots of this equation are the critical points. The solutions to this polynomial are:
[tex]\[ x_1 = \frac{5}{8} + \left( -\frac{1}{2} - \frac{\sqrt{3}i}{2} \right) \left( \frac{3\sqrt{7}}{64} + \frac{69}{512} \right)^{1/3} + \frac{9}{64 \left( -\frac{1}{2} - \frac{\sqrt{3}i}{2} \right) \left( \frac{3\sqrt{7}}{64} + \frac{69}{512} \right)^{1/3}} \][/tex]
[tex]\[ x_2 = \frac{5}{8} + \frac{9}{64 \left( -\frac{1}{2} + \frac{\sqrt{3}i}{2} \right) \left( \frac{3\sqrt{7}}{64} + \frac{69}{512} \right)^{1/3}} + \left( -\frac{1}{2} + \frac{\sqrt{3}i}{2} \right) \left( \frac{3\sqrt{7}}{64} + \frac{69}{512} \right)^{1/3} \][/tex]
[tex]\[ x_3 = \frac{9}{64 \left( \frac{3\sqrt{7}}{64} + \frac{69}{512} \right)^{1/3}} + \frac{5}{8} + \left( \frac{3\sqrt{7}}{64} + \frac{69}{512} \right)^{1/3} \][/tex]
3. Determine whether each critical point corresponds to a local maxima, minima, or neither:
To classify the critical points, we need to evaluate the second derivative [tex]\( g''(x) \)[/tex] at these points.
[tex]\[ g''(x) = \frac{d}{dx}(8x^3 - 15x^2 + 6x - 2) = 24x^2 - 30x + 6 \][/tex]
4. Evaluate [tex]\( g''(x) \)[/tex] at each critical point:
- At [tex]\( x_1 \)[/tex], [tex]\( g''(x_1) = -2.08248441322538 + 7.42836007216895i \)[/tex]
- At [tex]\( x_2 \)[/tex], [tex]\( g''(x_2) = -2.08248441322538 - 7.42836007216895i \)[/tex]
- At [tex]\( x_3 \)[/tex], [tex]\( g''(x_3) = 14.2899688264508 \)[/tex]
From these evaluations:
- Since [tex]\( g''(x_3) = 14.2899688264508 > 0 \)[/tex], [tex]\( x_3 \)[/tex] is a local minimum because a positive second derivative indicates that the function is concave up at this point.
- [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex] have complex values for their second derivatives, implying that these critical points are complex and thus not relevant for local minima or maxima in the real domain.
Hence, the polynomial function [tex]\( g(x) = 2x^4 - 5x^3 + 3x^2 - 2x + 1 \)[/tex] has a local minimum, not maxima.
Therefore, the correct answer is:
a. It has a local minima at approximately [tex]\( 1.65 \)[/tex].
