A question about the minimum/maximum of a convex function

  • Thread starter Thread starter pinodk
  • Start date Start date
  • Tags Tags
    Convex Function
pinodk
Messages
21
Reaction score
0
I would like to be sure in the following, not prove it, just have it confirmed...
If a function f is convex, then it has

1.) only one maximum and no minimum
2.) only one minimum and no maximum

infinity and -infinity are not included.
 
Physics news on Phys.org
It's not clear what you are asking. Since you say you want something "confirmed" it would appear you are making a statement: that either one of those two statements can be true for a convex function. That's not correct.

A convex function is a function f, such that the set {(x,y)| y> f(x) } is convex. Given that, on (2) is true.
(1) is true for a concave function.
 
Oh, ok, I didnt make a clear distinction between a convex and a concave function.

I didnt have a clear definition of convex and concave, so it makes more sense now, given your definition, and concave is then the opposite, and so makes (1) true.

thanks!
 
In particular, y= x2 is a convex function (the set of point above its graph is convex) and y= -x2 is a concave function (the set of points above its graph is concave).
 
pinodk said:
I would like to be sure in the following, not prove it, just have it confirmed...
If a function f is convex, then it has
1.) only one maximum and no minimum
2.) only one minimum and no maximum
infinity and -infinity are not included.

Convex...I knew it's concave up or concave down. Well, imagine the graph of y=x^2 or y=-x^2 depending on what you're talking about. It's a simple visualisation any Algebra 2 student can do...

bah...the guy above me wrote this exact thing.

However, I wonder, is there such a thing as Convex?
 
yes. In terms of functions: convex = concave up & concave = concave down.

I'm guessing that the whole up/down stuff is from a Calculus text by Stewart.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top