Carissa15
  • Carissa15
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Carissa15
  • Carissa15
Equation 1) \[f(x) =x^2 +\frac{ 1 }{ x }\] Equation 2) \[h(x) =x^3-9x^2+27x\]
zzr0ck3r
  • zzr0ck3r
Do you know how to take the derivative?
Carissa15
  • Carissa15
yes

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zzr0ck3r
  • zzr0ck3r
K what is the derivative of the first one?
Carissa15
  • Carissa15
2x -1x^-2 ?
zzr0ck3r
  • zzr0ck3r
right, now set that =0 and solve for x
Carissa15
  • Carissa15
not sure how to do that for the -x^-2 part? Do you add 1^2 to RHS?
Carissa15
  • Carissa15
to give\[x=\frac{ 1 }{ 2 }\]
zzr0ck3r
  • zzr0ck3r
woops sorry
zzr0ck3r
  • zzr0ck3r
\(2x-\frac{1}{x^2}=0\implies 2x^3-1=0\implies x=\frac{1}{\sqrt[3]{2}}\)
Carissa15
  • Carissa15
ok
Carissa15
  • Carissa15
so this gives the zero value.
zzr0ck3r
  • zzr0ck3r
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?
zzr0ck3r
  • zzr0ck3r
Finally we need to look at the zeros of the original equation and the right and left limits at 0.
zzr0ck3r
  • zzr0ck3r
Tie all the together and you get the graph.
Carissa15
  • Carissa15
I am with you
zzr0ck3r
  • zzr0ck3r
Also plug in \(\frac{1}{\sqrt[3]{2}}\) to see that it is the min for \(x>0\)
Carissa15
  • Carissa15
Ok, and we tell that the function is decreasing by comparing the values of each x=number that we are using?
zzr0ck3r
  • zzr0ck3r
we get |dw:1439178796606:dw|
zzr0ck3r
  • zzr0ck3r
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.
Carissa15
  • Carissa15
oh right ok.
zzr0ck3r
  • zzr0ck3r
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 \(((\dfrac{1}{\sqrt[3]{2}}), \infty)\)
zzr0ck3r
  • zzr0ck3r
left limit of 0 is -infty right limit is infty
zzr0ck3r
  • zzr0ck3r
Do the same sort of thing with the second one...
zzr0ck3r
  • zzr0ck3r
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.
Carissa15
  • Carissa15
so these 3 intervals then become the points of inflection? which then tell us the other important information.
zzr0ck3r
  • zzr0ck3r
points of inflection are when it changes from decreasing from increasing, that only happens once.
zzr0ck3r
  • zzr0ck3r
they are not intervals, they are points.
zzr0ck3r
  • zzr0ck3r
I am to lazy to evaluate that...
zzr0ck3r
  • zzr0ck3r
errrrr wait
Carissa15
  • Carissa15
That's ok. Thank you so much. What is the extrema?
Carissa15
  • Carissa15
is it infinity?
zzr0ck3r
  • zzr0ck3r
I just explained extrema, sorry inflection point is when it changes concavity
zzr0ck3r
  • zzr0ck3r
For that take the derivative of the derivative and find the zeros
zzr0ck3r
  • zzr0ck3r
if those zeros are defined in the original function, then they are inflection points
zzr0ck3r
  • zzr0ck3r
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
Carissa15
  • Carissa15
great thank you. This is my first time doing calculus and I am finding it difficult externally.
zzr0ck3r
  • zzr0ck3r
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.
Carissa15
  • Carissa15
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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