finding absolute extrema for f(x)=(sqrt 9-x^2) + x

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jessica716

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I need to find the absolute extrema for f(x)=(sqrt 9-x^2) + x on the interval [-3,0]. I figured out the 1st dervivative but i am having difficulity with the critical values and then finding the absolute max. and min.
 
Please reply showing your derivative and how far you have gotten in attempting to find the zeroes of the derivative, along with the values you obtained for the interval endpoints. (We can't error-check work we can't see! :oops: )

Please be complete. Thank you! :D

Eliz.
 
OK, f(x) =?(9-x²)+x, hence the domain of this function is -3?x?3.

dy/dx = [?(9-x²)-x]/?(9-x²) , Setting the derivative equal to zero, we have x =3/?2, x = -3/?2 is an extraneous root.
Ergo the absolute extrema is the max f(3/?2) = 6/?2 and the min f(-3)=-3.
 
Dr. Flim-Flam said:
Ergo the absolute extrema is the max f(3/?2) = 6/?2 and the min f(-3)=-3.
Click to expand...
Note: Critical points provide the locations of relative max/min points; the absolute max/min points are the highest/lowest points on the interval, which may or may not be at the endpoints and may or may not also be relative extrema.

Looking at the graph of this function, x = 3/sqrt[2] gives a relative max (the curve being higher here than at immediately adjacent points) and an absolute max (this being the location of the highest point of the curve within the entire inverval). And x = -3 gives an absolute min (not just "the min"), since it marks the lowest point of the graph on the interval.

Hope that helps! :D

Eliz.
 
Sorry about not replying with my work, my computer wasn't working for a while. The derivative i got is (-x/(sqrt9-x^2)) +1 and the critical values i got is x=1
 
jessica716 said:
The derivative i got is (-x/(sqrt9-x^2)) +1 and the critical values i got is x=1
Click to expand...
Um... You might want to check your work against the solution, provided earlier. :oops:

Eliz.
 
Is my derivative wrong? That is what my teacher wrote on the board for me. But I can't seem to figure out how to get the critical value x =3/?2, x = -3/?2. I have a graph on my paper and I'm pretty sure the answer is that there is no absolute extrema on the graph f(x)=(sqrt 9-x^2) + x on the interval [-3,0] and the points on the graph are (-3,-3) and (0,3), but i have to show my work and I'm not quite sure how to go about doing it. I know after I find the derivative I need to get the critical values
 
jessica716 said:
Is my derivative wrong?
Click to expand...
If you would review the replies you've received, you should note that, after some simplification and/or rearrangement, the derivative is correct. But how did you arrive at your critical point?

Also, you might want to try re-drawing your graph of f(x). There is very clearly a "hump" near the right-hand end, just as one would expect of the upper half of an origin-centered circle with radius r = 3, when added to an increasing linear function.

For us to figure out where you're going wrong, it would be very helpful if you showed your work. Thank you.

Eliz.
 
Thank you for helping me. I am sorry if I am not explaining where I am having problems correctly.
I am not sure if I will be able to really show my work, but I will try.

To get the critical values I know that i need to set the derivative equal to zero.
So I would have- (-x/(sqrt9-x^2)) +1=0. My teacher told me that whenever there is a numerator and a denominator to only use the numerator and set that equal to zero, because the denominator will always equal zero. I did that and got -x+1=0 which would then give me x=1. which is how I got the critical value of one.

Obviously this is wrong because I am supposed to get x =3/?2, x = -3/?2, as it was said above, but I can't figure out how to get this. I have been trying to follow what is written but i'm a bit confused.

Again, I appreciate the help.
 
Jessica:

dy/dx = -x/?(9-x²) + 1 = [-x+?(9-x²)]/?(9-x²)

Now,let -x+?(9-x²) = 0, what does x =?
 
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