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Determine all the critical points for the function f(x)= x^2 + (16/x)

Mathematics
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You found f' yet?
f '= 2x-(16/x^2)???
Yep! :)

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Other answers:

Now combine fractions! :)
\[f'=\frac{2x(x^2)-16}{x^2}\] right? :)
Thats what I got! :)
Now set Mr.top=0! :) and solve for x We would look at where the bottom is 0 except our function is not even define for x=0 So we are only look for when the top of f' is 0
\[2x(x^2)-16=0\]
Can you solve this for x?
x=2 ??
\[2x^3-16=0\] Yes that is the only real critical number \[x^3-8=0 \] \[(x-2)(x^2+2x+4)=0\] x=2 --- \[b^2-4ac=2^2-4(1)(4)=4-16=-12 <0 => x=2 \text{ is the only critical number}\] :) So good job! :)
Oh great! thank you
You deserve a medal! Seriously great job! :)

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