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Find the third degree polynomial Q such that Q(1)=1 Q'(1)=3 Q"(1)=6 and Q'''(1)=12​

User Trobol
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Answer:

A third degree polynomial, also known as a cubic polynomial, has the form of

Q(x) = ax^3 + bx^2 + cx + d

Given that Q(1)=1, Q'(1)=3, Q''(1)=6, and Q'''(1)=12, we can use these values to find the coefficients a, b, c, and d.

Q(1) = a(1)^3 + b(1)^2 + c(1) + d = 1

Q'(1) = 3a + 2b + c = 3

Q''(1) = 6a + 2b = 6

Q'''(1) = 6a = 12

Solving for a:

6a = 12

a = 2

Using the value of a, we can substitute it back into the other equations to find b, c and d.

Q''(1) = 6a + 2b = 6

6(2) + 2b = 6

12 + 2b = 6

2b = -6

b = -3

Q'(1) = 3a + 2b + c = 3

3(2) + 2(-3) + c = 3

6 - 6 + c = 3

c = 9

Q(1) = a(1)^3 + b(1)^2 + c(1) + d = 1

2(1)^3 + (-3)(1)^2 + 9(1) + d = 1

2 + (-3) + 9 + d = 1

8 + d = 1

d = -7

Therefore, the third degree polynomial Q that satisfies the given conditions is:

Q(x) = 2x^3 - 3x^2 + 9x - 7

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the polynomial Q(x) is Q(x) = x^3 + 2x^2 + x + 1

This polynomial satisfies the given conditions Q(1)=1, Q'(1)=3, Q"(1)=6, and Q'''(1)=12.

To check:

Q(1) = 1 + 2(1)^2 + 1(1) + 1 = 1

Q'(1) = 3 + 4 + 1 = 8

Q"(1) = 6

Q'''(1) = 6

It's important to note that the polynomial that I have provided may not be unique, as there may be other third degree polynomials that also satisfy the given conditions.

User Noah Mendoza
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