anonymous
  • anonymous
Find f. f ''(x) = 20x^3 + 12x^2 + 4 f(0) = 3 f(1) = 3 My Answer was x^5+x^4-2x+3 my answer was wrong, can someone tell me where I went wrong?
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
one sec
myininaya
  • myininaya
f'(x)=20x^4/4+12x^3/3+4x+C=5x^4+4x^3+4x+C f(x)=5x^5/5+4x^4/4+4x^2/2+Cx+D=x^5+x^4+2x^2+Cx+D
anonymous
  • anonymous
what math are you in Calc I or II

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anonymous
  • anonymous
1
anonymous
  • anonymous
okay I used integration, I'm assuming that you know how?
anonymous
  • anonymous
Yes
myininaya
  • myininaya
f(0)=D=3 f(1)=1+1+2+C+3=3 so C=-4
anonymous
  • anonymous
okay so take the integral of f double prime (c)
anonymous
  • anonymous
i mean (x)
anonymous
  • anonymous
so you get 20x^4/4 + 12x^3/3 + 4x
anonymous
  • anonymous
then simplify and do it once more
anonymous
  • anonymous
go it?
anonymous
  • anonymous
Medal pleaser ^.^
anonymous
  • anonymous
f(x)= x^5+x^4+C(x)+D
anonymous
  • anonymous
Plug f(0) in for x and set it equal to 3 makes D= 3 Plug f(1) in for x and set it equal to 3 makes C= -2 plugging those in was wrong
myininaya
  • myininaya
you are missing 2x^2 in f(x)
anonymous
  • anonymous
x^5 + x^4 + 2x^2 + c
anonymous
  • anonymous
myininaya solved it already
anonymous
  • anonymous
yeah
myininaya
  • myininaya
f(x)=x^5+x^4+2x^2+Cx+D
anonymous
  • anonymous
Thanks all

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