- #1
RedBarchetta
- 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|
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. 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. Thank you!
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