- 50

- 1

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|

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!

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: