Question

Determine the critical points, extremal points and intervals of monotony for the following functions on \(\mathbb{R}\):

a) \(f(x) = x^{4}+2x^{3}-2x^{2}+1\)

Answer 1

what's an interval of monotony?

Answer 2

hmm i dont know maybe its false translation, i ask a friend who can speak german and say you..

Answer 3

hang on... i'll google it...

Answer 4

ok..

Answer 5

it seems that an interval of monotony is just where the function is increasing/decreasing...

Answer 6

ok..

Answer 7

i thought it would be a false translation of my mentor but its ok i thin as you said

Answer 8

So then the question is asking for critical points, max/mins, and the intervals between the critical points.

Answer 9

and how we start ?

Answer 10

To find these, just take the derivative of the function. Can you do that yourself?

Answer 11

unfortunately not, but the year before last year i was able to

Answer 12

a friend of me tought me, and missed to pass this same exam with just 2 points :)

Answer 13

Just go back to the power rule I was describing earlier. So the derivative of \(x^n\) is \(nx^{n-1}\). Using this, the derivative of \(x^4\) is \(4x^3\). Using the same pattern what's the derivative of \(2x^3\)?

Answer 14

ok i remember

Answer 15

\[6x^{2}\]

Answer 16

is that?

Answer 17

Perfect. So now, what's the derivative of \[x^4+2x^3-2x^2+1\]

Answer 18

\[4x^{3}+6x^{2}-4x+1\]

Answer 19

can it be?

Answer 20

Just about. You should have\[4x^3+6x^2-4x\]Now that you have this, you need to find the critical points. To do this, set the derivative equal to 0, and solve for \(x\). That means we have\[4x^3+6x^2-4x=x(4x^2+6x-4)=0\]

Answer 21

If we continue factoring it, we get to \[2x(x+2)(2x-1)\]So this has zeroes at \[x=0,\quad x=-2,\quad x=\frac{1}{2}\]These are your critical points.

Answer 22

why we found here \[x(4x^2+6x-4)=0\]

Answer 23

i mean why we found this equation zero?

Answer 24

By definition, where the derivative is 0, is where the critical points are. We just looked at that because that's where the critical points are defined.

Answer 25

ok...

Answer 26

To find the extrema, plug the solutions of that back into the original equation. This would get you \[0^{4}+2(0)^{3}-2(0)^{2}+1=1\]\[(-2)^{4}+2(-2)^{3}-2(-2)^{2}+1=-7\]\[(1/2)^{4}+2(1/2)^{3}-2(1/2)^{2}+1=\frac{11}{16}\](I'm not entirely positive I did the math right on the last one)

Answer 27

These are the y-values of the function at the critical points. That means your extrema occur at the points \[(0, 1),\quad(-2, -7),\quad\left(\frac{1}{2}, \frac{11}{16}\right)\]

Answer 28

ok George, and is this our end solution ?

Answer 29

Now we just need to find the intervals of monotony. These are the intervals between the \(x\) values. So the intervals of monotony are \[(-\infty,-2),\quad(-2,0),\quad\left(0,\frac{1}{2}\right),\quad\left(\frac{1}{2},\infty\right)\]

Answer 30

ok..

Answer 31

that 11/16 should be 13/16...

Answer 32

I told you I wasn't sure about that :(

So wherever I wrote 11/16, replace it with 13/16

Answer 33

you should do a sign analysis to determine local max or local min

Answer 34

We're looking for the extremal points, so we need to find the points anyways.

Answer 35

when you solve for 0, those values are your critical points

Answer 36

have to determine critical points first then determine if they are extrema

Answer 37

2nd derivative will help identify points of inflection

Answer 38

Fair enough, but in this case they are all extrema.

Answer 39

ok thank you guys