# Thread: Local and Absolute Maximums/Minimums

1. ## Local and Absolute Maximums/Minimums

If given a local or absolute max/min value, can I use them to come up with the equation of an even or odd function using vertex form y= a(x-b)^n + c ??

2. ## Re: Local and Absolute Maximums/Minimums

I'm not sure what you mean by that. It is true that, if n is even, (x- b)^n is never negative so that a(x- b)^n is never negative if a is positive and never positive if a is negative. Then (x- b)^n+ c is
1) equal to c at x= b and, if a>0, larger than c for x not equal to b.
2) equal to c at x= b and, if a< 0, less than c for x not equal to b.

In the first case, the graph goes down to a "vertex" at (a, c) and back up again.
In the second case, the graph goes up to a "vertex" at (a, c) and back down again.

3. ## Re: Local and Absolute Maximums/Minimums

Originally Posted by sakonpure6
If given a local or absolute max/min value, can I use them to come up with the equation of an even or odd function using vertex form y= a(x-b)^n + c ??
The equation you wrote y= a(x-b)^n + c must have n zeros. if all of them are real and without multiplicity, then you will have n-1 local extrema. Also, if the domain is R, then it will have two absolute extrema if if n<0, one absolute extrema.

4. ## Re: Local and Absolute Maximums/Minimums

Originally Posted by votan
The equation you wrote y= a(x-b)^n + c must have n zeros. if all of them are real and without multiplicity, then you will have n-1 local extrema. Also, if the domain is R, then it will have two absolute extrema if if n<0, one absolute extrema.
But "if all of them are real and without multiplicity" does not occure here. The zeros are obviously x= b+ (-c/a)^{1/n}. If n is odd, that has exactly one real root. If n is even and -c/a is positive, there are exactly two real roots. Most of what you say here is irrelevant to the question asked.

(And, once again, I am not clear whether or not you are doing this intentionally!)