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Equation 1) \[f(x) =x^2 +\frac{ 1 }{ x }\]
Equation 2) \[h(x) =x^3-9x^2+27x\]

Do you know how to take the derivative?

yes

K what is the derivative of the first one?

2x -1x^-2 ?

right, now set that =0 and solve for x

not sure how to do that for the -x^-2 part? Do you add 1^2 to RHS?

to give\[x=\frac{ 1 }{ 2 }\]

woops sorry

\(2x-\frac{1}{x^2}=0\implies 2x^3-1=0\implies x=\frac{1}{\sqrt[3]{2}}\)

ok

so this gives the zero value.

Finally we need to look at the zeros of the original equation and the right and left limits at 0.

Tie all the together and you get the graph.

I am with you

Also plug in \(\frac{1}{\sqrt[3]{2}}\) to see that it is the min for \(x>0\)

we get
|dw:1439178796606:dw|

oh right ok.

left limit of 0 is -infty
right limit is infty

Do the same sort of thing with the second one...

points of inflection are when it changes from decreasing from increasing, that only happens once.

they are not intervals, they are points.

I am to lazy to evaluate that...

errrrr wait

That's ok. Thank you so much. What is the extrema?

is it infinity?

I just explained extrema, sorry
inflection point is when it changes concavity

For that take the derivative of the derivative and find the zeros

if those zeros are defined in the original function, then they are inflection points

great thank you. This is my first time doing calculus and I am finding it difficult externally.