Question

I found the derivative of a functino and now I need help finding the critical/stationary values. (click to see)

Answer 1

\[ f'(x)=\frac{ x^2+2x-1 }{ x^2+2x+1 }\]

Answer 2

Bottom can be factored to find vertical asymptotes. Top looks to require quadratic formula to find zeroes.

Answer 3

ok thanks

Answer 4

If you want stationary inflection points, you'll need to find where the second derivative =0.

Answer 5

I just needed to find the abs max/min

Answer 6

I found that there is only an absolute max within the interval I am trying to find. The x-value for the abs max is -2-sqrt2

Answer 7

How do I find an exact value for the y-coordinate of the abs max?
Where the original function is...

Answer 8

\[ f(x)=\frac{ x^2+1}{ x+1}\]

Answer 9

If you have the original function, plug-and-chug.

Answer 10

I have found a decimal approximation but I do not know how to obtain the exact value for y.

Answer 11

Hmm. Gimme a sec to work it out, I'll let you know what I see.

Answer 12

What is the interval you're looking in?

Answer 13

(-5,-1)

Answer 14

I'm getting \(-1-\sqrt{2}\) for the critical point.

Answer 15

ya i just checked the book, and they have the same value...

Answer 16

i just realized i got -2-sqrt because i only divided 2 from the coefficient of sqrt2 and not the integer -2

Answer 17

If you want the exact y value, then you need to put the exact x value in the function.

\[f(-1-\sqrt{2})=\frac{(-1-\sqrt{2})^2+1}{-\sqrt{2}}\]

Answer 18

Also, you'll either need to look for intervals of increase and decrease, or put the critical point into the second derivative to test for concavity to determine rel. min. or max.

Answer 19

Yes, I used a number line and found the values for f'(x)

Answer 20

Ok, so to verify absolute extrema, you also need to check the endpoints of the interval.

Answer 21

Even if it is an open interval?

Answer 22

@CliffSedge

Answer 23

I'm pretty sure, yeah. Even if it's open, you can find the limit of the function at those points.

Answer 24

From Wolframalpha: "An open interval is an interval that does not include its end-points."

Answer 25

There is that asymptote at -1, though.

Answer 26

Right, the function is not defined at the endpoints, but the function still exists in that neighborhood.

Answer 27

in other words. You wouldn't bother checking -5, but you could check -4.99999999999

Answer 28

That said, to find the min/max near the end-points I believe I would have to choose a value sufficiently close to -5 and -1 but where and how would I determine what is close enough?

Answer 29

Use properties of limits. You should be able to use -5 as is no problem.
-1 is a discontinuity though, so you'll need something like L'Hopital's rule to find the limit there.

Answer 30

Ok, well thank you for the help. I will find out if I need to find the limits close to the ends of an open interval from my teacher just to be sure. Thanks a lot for the help...

Answer 31

excuse me, not the limits, the max/min values...

Answer 32

When in doubt, graph it.