A community for students.
Here's the question you clicked on:
 0 viewing
Carissa15
 one year ago
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
Carissa15
 one year ago
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

This Question is Closed

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0Equation 1) \[f(x) =x^2 +\frac{ 1 }{ x }\] Equation 2) \[h(x) =x^39x^2+27x\]

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1Do you know how to take the derivative?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1K what is the derivative of the first one?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1right, now set that =0 and solve for x

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0not sure how to do that for the x^2 part? Do you add 1^2 to RHS?

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0to give\[x=\frac{ 1 }{ 2 }\]

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1\(2x\frac{1}{x^2}=0\implies 2x^31=0\implies x=\frac{1}{\sqrt[3]{2}}\)

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0so this gives the zero value.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1Ok, 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
 one year ago
Best ResponseYou've already chosen the best response.1Finally we need to look at the zeros of the original equation and the right and left limits at 0.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1Tie all the together and you get the graph.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1Also plug in \(\frac{1}{\sqrt[3]{2}}\) to see that it is the min for \(x>0\)

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0Ok, and we tell that the function is decreasing by comparing the values of each x=number that we are using?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1we get dw:1439178796606:dw

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1we 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.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1After 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
 one year ago
Best ResponseYou've already chosen the best response.1left limit of 0 is infty right limit is infty

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1Do the same sort of thing with the second one...

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1up 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
 one year ago
Best ResponseYou've already chosen the best response.0so these 3 intervals then become the points of inflection? which then tell us the other important information.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1points of inflection are when it changes from decreasing from increasing, that only happens once.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1they are not intervals, they are points.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1I am to lazy to evaluate that...

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0That's ok. Thank you so much. What is the extrema?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1I just explained extrema, sorry inflection point is when it changes concavity

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1For that take the derivative of the derivative and find the zeros

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1if those zeros are defined in the original function, then they are inflection points

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1This 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
 one year ago
Best ResponseYou've already chosen the best response.0great thank you. This is my first time doing calculus and I am finding it difficult externally.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.0Awesome, 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 :)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.