# Math challenge

1. Nov 1, 2005

### JustinJS

ok here is a good one i found
a) f'(-1)=f'(1)=0
b) f'(x)<0 when -1<x<1
c) f'(x)>0 when x<-1 and x>1
d) f(-1)=4; f(1)=0; and f(0) has no value
e) f''(x)<0 when x<0 and f''(x)>0 when x>0
what is the original f?

2. Nov 2, 2005

### Haelfix

Are you sure f(-1) = 4 and not f(-1) = -4?

3. Nov 2, 2005

4. Nov 2, 2005

### Galileo

Does this even have a unique solution? You've mostly given inequalities, so I suspect not.

Let's look at the interval (<,-1]
f'(-1)=0, the graph is horizontal at x=-1
f'(x)>0 when x<-1. It is increasing all the way to x=-1
f(-1)=4, goes through the point (-1,4)
e) f''(x)<0 when x<0. And is concave.

Well, I can think of one function in this region which satisfies it. A mountain parabola like -(x-a)^2+b. The vertex should be at -1, so that gives a. It goes through (-1,4) so this gives b.
Surely you can find more functions. It doesn't even matter whether we 'compress' the graph to the line x=-1, so -k(x-a)^2+b works just as well for any k>0.

5. Nov 2, 2005

### JustinJS

idk it was on a quiz in (AP calc) i had the other day and i got it wrong and was wonder if anyone could get an answer for it.

6. Nov 2, 2005

### Haelfix

I guessed the following solution

F(x) = c1x^-1 + c2 x + c3

You can check it all works out, minus one problem (as I mentioned above) the condition f(-1) =4 clashes with the concavity conditions on f''.. C1 needs to be strictly positive.

Nor would it be unique, the inverse power term works so long as its power is odd.

so try something like

f = x^-1 +x -2