• Support PF! Buy your school textbooks, materials and every day products Here!

Logarithmic Antiderivative

Find f. (x>0)
f''(x) = x^(-2)
x > 0
f(1) = 0
f(8) = 0

Alright, everything was fine until one point....okay. First derivative:
f''(x)=x[tex]^{-2}[/tex]
f'(x)=-x[tex]^{-1}[/tex]+C

Now here's where i'm not sure: in the next step, it will be x raised to zero, which is one. times 1/0, which is zero, but...my physics professor informed me that it is actually, the natural log of the absolute value of x.

f(x)=-ln|x|+Cx+D

Now, use the above input & output values of f(x) to divulge the constants.

f(1)=C+D=0
f(8)=-ln|8|+8c+D=0

C+D=0
8c+D=ln|8|
8c-c=ln|8|
7c=ln|8|

C=[tex]\frac{ln|8|}{7}[/tex]


Now, plug in:

f(x)=-ln|x|+[tex]\frac{x*ln|8|}{7}[/tex]-[tex]\frac{ln|8|}{7}[/tex]

Does this look right? I'm not quite sure about the x in the second term in that equation. I only have one more try on my online homework thing. :rofl: Thank you!
 
Last edited:

Answers and Replies

Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,598
6
Looks good to me :approve:
 
HallsofIvy
Science Advisor
Homework Helper
41,734
893
Since 8 is a positive number, |8|= 8. Your "online homework thing" may object to |8| rather than 8. (That's why I hate those things!)
 
Last edited by a moderator:
Thank you! The answer was correct, and the checker decided to take my answer. What a surprise.
 

Related Threads for: Logarithmic Antiderivative

  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
15
Views
964
  • Last Post
Replies
7
Views
548
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
3
Views
5K
Top