# Single Variable function, local max/min is absolute max/min

1. Oct 10, 2009

### calorimetry

1. The problem statement, all variables and given/known data
Consider a function of one variable, f(x), which is continuous on an interval I. If f has a single critical point, x = a, in I, then if that point is a local maximum (or minimum) it must also be an absolute maximum. Explain why this is.

2. Relevant equations
Intermediate Value Theorem
Rolle's Theorem

3. The attempt at a solution
Critical point x=a, in I, is a local maximum or minimum, thus f'(x)=0 at x=a.

The function f thus must either concave up (for a to be a local min) or concave down (for a to be a local max).

Since both cases are similar, let consider x=a is a local max.

For a local max, f'(x) is decreasing to 0 as x approaches a from the left and f'(x) is increasing negative away from 0 as x deviate away from a on the right.

Since there is only one single critical point at x=a, the concavity of the graph will not change. f(x) on the left of is an increasing function to f(a) and f(x) on the right of f is a decreasing function away from f(a). Thus x=a must also be the absolute maximum.

Is my reasoning correct? Or did I unconsciously assumed something that I shouldn't have?

2. Oct 10, 2009

### Donaldos

Why don't you use the theorems you mentioned?

Assume there exists $$x_0$$ such that $$f(x_o)>f(a)$$.

What does the intermediate value theorem tell you?

3. Oct 10, 2009

### calorimetry

If f(x_o) > f(a), then according to the intermediate value theorem, there should be another value in the interval I where f(x) = f(a).

Then applying Rolle's theorem, there must be a point between x and a such that the derivative of this point is 0, hence there must be a second critical point, which contradicts with single critical point assumption.

Thanks for the help, I didn't consider proving it using proof of contradiction. Even if I did, I would not know where to begin :d. Thanks again.

Last edited: Oct 10, 2009