What is the Antiderivative of x^(-2) with Given Conditions?

[RedBarchetta]

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$$^{-2}$$
f'(x)=-x$$^{-1}$$+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=$$\frac{ln|8|}{7}$$


Now, plug in:

f(x)=-ln|x|+$$\frac{x*ln|8|}{7}$$-$$\frac{ln|8|}{7}$$

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. Thank you!

 

[Hootenanny]

Looks good to me

 

[HallsofIvy]

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!)

 

[RedBarchetta]

Thank you! The answer was correct, and the checker decided to take my answer. What a surprise.