Question

Find a polynomial equation of degree 3 such that f(0)= 31, f'(1)= 4, f"(2)= 2, and f'"(3)=6

Answer 1

start with f'''(x).

if f'''(3) = 6, then for simplicity, we could assign f'''(x) = 2x.

the integral of f'''(x) is f''(x) + C.

f''(x) = x^2 + C.

let's plug in 2, and adjust C such that f''(2) = 2.

f''(2) = 2^2 + C = 2. ---> C = -2

so f''(x) = x^2 - 2.

is this enough info to do the rest or should i keep going?

Answer 2

so is the next step x^2-2=2? i don't know how to the next step

Answer 3

sorry.

don't do this. it'll lead to a 4th order polynomial.

Answer 4

assign f'''(x) = 6. therefore f'''(3) = 6 holds true.

the integral of 6 is 6x + C = f''(x)

f''(2) = 2.

6(2) + C = 2

C = -10

so f''(x) = 6x -10.

now to find f'(x), we take the integral of f''(x). do you know what that is?

Answer 5

x^6-10x+c?

Answer 6

close. it's 3x^2 - 10x + C

\[\int\limits_{}^{} x^n = \frac{ x^{n+1} }{ n+1 }\]
\[\int\limits_{}^{}6x = 6\int\limits_{}^{}x = 6*\frac{ x^2 }{ 2 } = 3x^2\]

Answer 7

\[\int\limits_{}^{}x^4 = \frac{ x^5 }{ 5 }\]

Answer 8

i didn't even consider, have you learned integrals yet?

Answer 9

there's a longer way to do it with only derivatives

Answer 10

no we haven't learned integrals yet haha i was wondering why this made zero sense to me

Answer 11

lol :P

Answer 12

so a 3rd order polynomial looks like this:\[f(x) = ax^3 + bx^2 + cx + d\]
the derivative of that, 2nd order poly, would look like this:\[f'(x) = 3ax^2 + 2bx + c\]and\[f''(x) = 6ax + 2b\]\[f'''(x) = 6a\]
given our info:
f'''(3) = 6

we an plug 3 in our "general" f'''(x) to find out what a is

6 = 6a

a = 1. this is true for all of the equations derived from each other.

given: f''(2) = 2, a = 1 and f''(x) = 6ax + 2b

you can figure out b

2 = 6(1)(2) + 2b ----> b = -5

and then you can find c, then d

Answer 13

i'll start the next one:

given f'(1) = 4:\[4 = 3(1)(1^2) + 2(-5)(1) + c\]
solve for c

Answer 14

so c= 11

Answer 15

and then f(x)= x^3 -5x^2 +B

Answer 16

and d= 31?

Answer 17

yes

Answer 18

okay i got it :) thank you so much!!

Answer 19

glad i could help :)