IDNLearn.com offers a unique blend of expert answers and community-driven insights. Our experts are ready to provide in-depth answers and practical solutions to any questions you may have.
Sagot :
To determine the multiplicity of the root [tex]\( x = 2 \)[/tex] for the polynomial [tex]\( y = x^3 - 3x^2 + 4 \)[/tex], we follow these steps:
1. Define the Polynomial:
[tex]\[ P(x) = x^3 - 3x^2 + 4 \][/tex]
2. Evaluate the Polynomial at [tex]\( x = 2 \)[/tex]:
[tex]\[ P(2) = (2)^3 - 3(2)^2 + 4 = 8 - 12 + 4 = 0 \][/tex]
Since [tex]\( P(2) = 0 \)[/tex], [tex]\( x = 2 \)[/tex] is indeed a root of the polynomial.
3. Find the First Derivative:
[tex]\[ P'(x) = 3x^2 - 6x \][/tex]
Evaluate the first derivative at [tex]\( x = 2 \)[/tex]:
[tex]\[ P'(2) = 3(2)^2 - 6(2) = 12 - 12 = 0 \][/tex]
4. Find the Second Derivative:
[tex]\[ P''(x) = 6x - 6 \][/tex]
Evaluate the second derivative at [tex]\( x = 2 \)[/tex]:
[tex]\[ P''(2) = 6(2) - 6 = 12 - 6 = 6 \][/tex]
Since [tex]\( P''(2) \neq 0 \)[/tex], this means that the second derivative is non-zero at [tex]\( x = 2 \)[/tex].
To summarize:
- [tex]\( P(2) = 0 \)[/tex]
- [tex]\( P'(2) = 0 \)[/tex]
- [tex]\( P''(2) \neq 0 \)[/tex]
These conditions imply that [tex]\( x = 2 \)[/tex] is a root with multiplicity 2, because the polynomial and its first derivative are zero at [tex]\( x = 2 \)[/tex], but the second derivative is non-zero. Therefore, the root [tex]\( x = 2 \)[/tex] has a multiplicity of [tex]\( 2 \)[/tex].
So, the correct answer is:
2
1. Define the Polynomial:
[tex]\[ P(x) = x^3 - 3x^2 + 4 \][/tex]
2. Evaluate the Polynomial at [tex]\( x = 2 \)[/tex]:
[tex]\[ P(2) = (2)^3 - 3(2)^2 + 4 = 8 - 12 + 4 = 0 \][/tex]
Since [tex]\( P(2) = 0 \)[/tex], [tex]\( x = 2 \)[/tex] is indeed a root of the polynomial.
3. Find the First Derivative:
[tex]\[ P'(x) = 3x^2 - 6x \][/tex]
Evaluate the first derivative at [tex]\( x = 2 \)[/tex]:
[tex]\[ P'(2) = 3(2)^2 - 6(2) = 12 - 12 = 0 \][/tex]
4. Find the Second Derivative:
[tex]\[ P''(x) = 6x - 6 \][/tex]
Evaluate the second derivative at [tex]\( x = 2 \)[/tex]:
[tex]\[ P''(2) = 6(2) - 6 = 12 - 6 = 6 \][/tex]
Since [tex]\( P''(2) \neq 0 \)[/tex], this means that the second derivative is non-zero at [tex]\( x = 2 \)[/tex].
To summarize:
- [tex]\( P(2) = 0 \)[/tex]
- [tex]\( P'(2) = 0 \)[/tex]
- [tex]\( P''(2) \neq 0 \)[/tex]
These conditions imply that [tex]\( x = 2 \)[/tex] is a root with multiplicity 2, because the polynomial and its first derivative are zero at [tex]\( x = 2 \)[/tex], but the second derivative is non-zero. Therefore, the root [tex]\( x = 2 \)[/tex] has a multiplicity of [tex]\( 2 \)[/tex].
So, the correct answer is:
2
We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. Your search for solutions ends here at IDNLearn.com. Thank you for visiting, and come back soon for more helpful information.