Question

Calculus 1: Concavity
f(x) = x^4 - 4x^2 + (7/4)
f'(x) = 4x(x^2 - 2). Critical Points: 0, +/- sqrt(2)
f''(x) = 12x^2 - 8

Answer 1

What do you need to know?

Answer 2

up or down I am guessing?

Answer 3

The intervals where the function is concave up and down. I think I'm supposed to solve f''(x) >0 and f''(x) < 0 individually, right? Then define the intervals using those>

Answer 4

You could do that. You could also solve for the points of inflection and use the end points to tell you where it is concave up and concave down. Know what I mean?

Answer 5

Solve for the zeros of the second derivative and you will know where you points of inflection are on your first function right?

Answer 6

your**

Answer 7

Yes.
For 12x^2 - 8 < 0, \[-\sqrt{2/3} < x < \sqrt{2/3}\]
For 12x^2 - 8 > 0, \[x > \sqrt{2/3}\]. This is kind of where I got stuck though. How do I determine the intervals of concavity for f''(x) < 0?

Answer 8

Okay so you have the inflection points. What is the degree of your original function?

Answer 9

it is a 4th-degree function

Answer 10

So how many curves would you expect to find in the graph?

Answer 11

two?

Answer 12

Are you sure?

Answer 13

first degree is just a line a second is a parabola a third has two curves and a fourth has?

Answer 14

wolfram shows a parabola, but my TI-83 shows a plot with 3

Answer 15

Correct you should expect 3 curves, but this isn't always the case. You need to find the multiplicity of the first derivative understand?

Answer 16

your TI-83 is just showing the graph at a different scale so it looks like there is 1 but there is actually 3; however, we are going to pretend we don't know how many curves there are so you will understand better how to do this. Trust me it will probably be useful on an exam.

Answer 17

Sorry I mean wolfram is showing it a different scale.

Answer 18

You're kind of getting off into a tangent besides the original issue I had, what do I do with f''(x)<0 to determine concativity?

Answer 19

No I am not haha listen if you just follow the steps it is easy all you need to do is factor the first derivative can you do that?

Answer 20

I guarantee you can =)

Answer 21

How is 4x(x^2 -2) not the first derivative already factored?

Answer 22

That is why I said I guarantee you can =)

Answer 23

Asking question and answering questions is the best way to learn even if the seem pointless

Answer 24

\[1/4(4x ^{4} + 4x ^{2} + 7)\]

Answer 25

Okay could you factor the inside? Use wolfram is you want I just want you to get the general method.

Answer 26

if**

Answer 27

you're look for something with this (a+b)^n know what I mean?

Answer 28

You just need to know whether the exponents on the outside of the brackets is odd or even and for what zero, or is it for all zeros?

Answer 29

you know I appreciate your willingness to go in-depth but I'd really just like help with the original problem

Answer 30

Alright I appreciate your honesty, but trust me you will need this eventually. It will be useful and fast once you know how to do it, and it is your original problem.

Answer 31

then?

Answer 32

Sorry forget everything I just said it was supposed to be the first derivative anyways the first derivative has multiplicity of 1 so it is odd this means for the zeros the first derivative goes through the x-axis which means the first function changes signs so if you know the end behaviours you can just write down the intervals around the maxima and minima of you first function then do a quick sketch and you will see where the graph is increasing/ decreasing and what its concavity at each minima is

Answer 33

minima/maxima

Answer 34

If you know the multiplicity of the first derivative, the degree, whether the function is odd or even, the maxima/minima then you can write sketch the graph really quick and you know the concavities.

Answer 35

thanks for sticking it out with me. but I would like to go in order as the instructor did, which means I wait until I have all information to graph. so how would I find the intervals of concavity from my work above?

Answer 36

Does what I just said help clarify your questions? If not I will try to give a better, but still short list of how to solve the problem.

Answer 37

Okay to put it really easy lets pretend all graphs will work out so that the degree-1 will tell us how many maxima/minima then all we need is the end behaviours and the locations of the maxima/minima in order to figure out where the graph is concave up and concave down. Does this make sense?

Answer 38

you know we're not going anywhere with this so unless you can help me get what I want from the information I supplied, then I'm closing the question and doing it myself. I'll give you a medal for trying but this is not working

Answer 39

Okay just relax I want to help you it is really easy. We are like 2 seconds from being done. You just need to know how does the graph start and end? does it start of positive or negative? does it end positive or negative?

Answer 40

the graph doesn't end, it's continuous, if you're asking me where it increases and decreases then that information I know

Answer 41

lol I know it goes on forever I mean the end behaviours do you know this like the ends go off infinitely either positive or negative direction are they decreasing or increasing at the ends? By ends I mean after the curves.

Answer 42

those lines are the end behaviours are they increasing in the y or decreasing?

Answer 43

decreasing then increasing

Answer 44

okay so imagine you draw in your 3 maxima/minima. You know the you have 3 of them because it is a 4th degree function, and the multiplicity of the first derivative is odd. mark down the zeros in your mind and imagine drawing the function. where is the function concave up/down here I will get you started you draw what makes sense to you back.