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OpenStudy (anonymous):
take the derivative, set it equal to zero and solve
OpenStudy (anonymous):
can you walk me through finding the derivative
OpenStudy (anonymous):
piece by piece
what is the derivative of \(x\) ?
OpenStudy (anonymous):
1
OpenStudy (anonymous):
k
and what is the derivative of \(\frac{1}{x}\) ?
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OpenStudy (anonymous):
x
OpenStudy (anonymous):
oh no!
OpenStudy (anonymous):
thats where im mistaken at
OpenStudy (anonymous):
the derivative of \(\frac{1}{2}x^2\) is \(x\)
OpenStudy (anonymous):
oh ok
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OpenStudy (anonymous):
the derivative of \(\frac{1}{x}\) is \(-\frac{1}{x^2}\)
OpenStudy (anonymous):
so 1/x is?
OpenStudy (anonymous):
ok so you set -1/x^2 to zero and solve
OpenStudy (anonymous):
ok im lost in sovling this
OpenStudy (anonymous):
ok no I use a number under 0 and a number over right
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OpenStudy (anonymous):
oh no, set
\[1-\frac{1}{x^2}=0\]
OpenStudy (anonymous):
forgot about the 1
OpenStudy (anonymous):
\[f(x)=1+\frac{1}{x}\]
\[f'(x)=1-\frac{1}{x^2}\] set
\[f'(x)=0\] get
\[1-\frac{1}{x^2}=0\] solve for \(x\)
OpenStudy (anonymous):
you can solve this in your head with no algebra, or you can do algebra if you like
OpenStudy (anonymous):
x^2=1
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OpenStudy (anonymous):
right
OpenStudy (anonymous):
x=plus or minus 1
OpenStudy (anonymous):
right
OpenStudy (anonymous):
so its critical numbers would be -1,1,0
OpenStudy (anonymous):
also \(0\) might be a critical point, but 0 is not in the domain of the original function, so it would be a miracle if it was in the domain of the derivative
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OpenStudy (anonymous):
so i would not call it a "critical point" since it is not in the domain of \(f\)
your math teacher might disagree with me