# Thread: max and min urgent help

1. ## max and min urgent help

1)consider f(x)= 2- x for x<=0 and k-x^2 for x>0. For which values of k belonging to R has f(x) got a point of local minimum at x=0?
2) If y= f(x) has a point of global minimum at x=6, which of the following functions has certainly got a point of global max at x=0? The possible answers are a) y= -6f(x); b)y=-f(6x); c)y=f(x+6); d)y= -f(x+6) I feel like crying...though they're stupid and basical things...help me please..

2. Originally Posted by Aglaia
consider f(x)= 2- x for x<=0 and k-x^2 for x>0. For which values of k belonging to R has f(x) got a point of local minimum at x=0?
For x = 0 to be a local minimum we need to have f(0) the smallest value in an open neighborhood about x = 0. Now, f(0) = 2, and this is smaller than any f(x) with x<0. So far, so good. Now we need to arrange a k value such that f(x) for x near 0 is greater than 2. Thus we need k>2.

-Dan

3. Originally Posted by Aglaia
If y= f(x) has a point of global minimum at x=6, which of the following functions has certainly got a point of global max at x=0? The possible answers are a) y= -6f(x); b)y=-f(6x); c)y=f(x+6); d)y= -f(x+6)
Since f(x) has a minimum at x = 6 we automatically know that -f(x) has a maximum at x = 6. Thus answer a) is correct.

b) y = -f(6x) would have a max at 6x = 6 or x = 1.

c) y = f(x+6) would have a min at x+6 = 6 or x = 0. We have no information about a max point.

d) y = -f(x+ 6) would have max at x+6 = 6 or x = 0.

-Dan

4. Originally Posted by Aglaia
1)consider f(x)= 2- x for x<=0 and k-x^2 for x>0. For which values of k belonging to R has f(x) got a point of local minimum at x=0?
Use geometric intution.

The curve y=2-x x<=0 is a line with negative slope at (0,2)

The curve y=k-x^2 is a entire set of half shaped upside down parabolas.

We want a relative minimum at x=0.
Meaning some open interval containing x as a minimum point.
Since the parabola is always going down we cannot have k<=2 because otherwise the right side points of x=0 will be above the parabola. So we need k>2, that will make the parabola on above the line ensuring there is some open interval.
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So what you posted before me, I explained better.

5. well i meant zero not six...i wrote what is written on the book "point of global max at x=0" not x=6..so, what's the answer?

6. Originally Posted by Aglaia
well i meant zero not six...i wrote what is written on the book "point of global max at x=0" not x=6..so, what's the answer?
This is a problem since, following the logic I used in my solution both -6f(x) and -f(6x) would have a max at x = 0. So your answer would be both a) and b). If your problem is such that this is allowed then this is your answer.

-Dan

7. well it's a multiple choice question and i'm allowded to answer ticking just one...i'm getting down...by the way if you say -f(x) of course you' re turning the function upside down..dut then it was written" -6f(x)" which would probably change the things provided that there has been a prnting mistake or smth like that...what do you think?

8. Originally Posted by Aglaia
well it's a multiple choice question and i'm allowded to answer ticking just one...i'm getting down...by the way if you say -f(x) of course you' re turning the function upside down..dut then it was written" -6f(x)" which would probably change the things provided that there has been a prnting mistake or smth like that...what do you think?
-f(x) turns the function upside-down (technically called a "reflection over the x-axis).

Answer b) was -f(6x). -f(6x) turns the function upside-down, but also "shrinks" the x-axis by a factor of 6. However this stretching won't affect the point (0, f(0)) because x = 0 is the center of the shrinking, ie it doesn't move. So we would still have a maximum at x = 0. (The function -6f(x) flips the function upside-down and "stretches" the y-axis by a factor of 6. This would still leave x = 0 as a maximum.) Either way b) is still an acceptable answer.

There could be a typo, I'm just not sure what the typo would have been. You're going to have to take this one to your teacher I think.

-Dan

9. oh my God...i definitively need to talk to my prof. anyway thank you..i feel better as i was sure i was so stupid because those things seemed to be very easy and i couldn't come up with any sort of solution...they are not so easy..on the contrary...thanks a lot!

10. Originally Posted by Aglaia
oh my God...i definitively need to talk to my prof. anyway thank you..i feel better as i was sure i was so stupid because those things seemed to be very easy and i couldn't come up with any sort of solution...they are not so easy..on the contrary...thanks a lot!
Actually they aren't that bad. You just need some time to get used to them. Good luck with your professor!

-Dan