I need to find the intervals on which the following equations have zeroes, extrema, points of inflection and intervals increasing or decreasing, as well as concave up or down. I need to show these on a graph but is there a better way to present the data (before graphing) instead of just x=1 and sub into equation, then x=2, and so on? Thanks in advance

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I need to find the intervals on which the following equations have zeroes, extrema, points of inflection and intervals increasing or decreasing, as well as concave up or down. I need to show these on a graph but is there a better way to present the data (before graphing) instead of just x=1 and sub into equation, then x=2, and so on? Thanks in advance

Mathematics
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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

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Other answers:

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.
Ok, so we have some points to look at. First we have \(0\) will give an asymptote so we look to the left of \(0\) We can put in any number to the left of \(0\) in the derivative and get a negative number so we know the original equation is decreasing on the left of \(0\). So now we at the derivative between \(0\) and \(\frac{1}{\sqrt[3]{2}}\) and get that it is positive so the original equation is also decreasin on \((0, \frac{1}{\sqrt[3]{2}})\). Do the same thing to the right of the fraction solution to get its decreasing. you with me?
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 \(x>0\)
Ok, and we tell that the function is decreasing by comparing the values of each x=number that we are using?
we get |dw:1439178796606:dw|
we plug in a number like -3 into the derivative, and that gives a negative values, so the rate of change is negative, so the original function is decreasing.
oh right ok.
After knowing we cant have x=0 in the original equation, and seeing the derivative only had one zero, we know we only need to look in 3 intervals \((-\infty, 0),(0, \dfrac{1}{\sqrt[3]{2}}),\) and \(((\dfrac{1}{\sqrt[3]{2}}), \infty)\)
left limit of 0 is -infty right limit is infty
Do the same sort of thing with the second one...
up until calc you put in the y int and the 0's and used some other little tools to figure out what some limited graphs look like, you are doing the same thing here but with most graphs, and a much better idea of what the graph looks like.
so these 3 intervals then become the points of inflection? which then tell us the other important information.
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
This I dont think has any zeros and so even though the graph is concave down for x< 0 and concave up for x>0 we never actually change concavity ....because of the asymptote
great thank you. This is my first time doing calculus and I am finding it difficult externally.
I bet, we did plenty here because I thought this was not so new to you. For your question, the easiest way would be to 1) find the zeros of the derivative and notice the asymptote 2) look at the intervals on the left and right of those zeros to determine behavior to find extrema 3) do the same thing for the second derivative to find inflection points.
Awesome, thank you :) I seem to spend hours trying to figure things out on google or khan academy. Thank you so much for your help, I should be fine with the second equation now that I know the steps :)

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