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Mathematics 24 Online
OpenStudy (anonymous):

find all critical numbers of f(x)=x+(1/x)

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 \(\frac{1}{x}\) ?

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

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

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

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

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

OpenStudy (anonymous):

the hint said it had 3

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