Find the minimum

[George Saliaris]

Let f(x) = e^(x^2 - x) + e^(1-x).Prove that the minimum value of f(x) is 2 (without the use of derivatives).... What I have tried:
1) a wide range of "brute force" methods
2) completing the "square" 3) use of some basic inequalities (like x + 1/x >=2... All of these did not get me somewhere.. Any hints?

Edit : Using Bernoulli's inequality I did it but I doubt our teacher wants us to do that since we have not learned this inequality

 

[tkhunny]

[math]e^{x^{2}-x} + e^{1-x} = \left(e^{x}\right)^{x-1} + \left(e^{-1}\right)^{x-1}[/math]

 

[Dr.Peterson]

A key may be that you are not told to find the minimum, but to prove a given minimum. That means you can use a different kind of strategy.

For what value of x does f(x) = 2? It's easy (at least to find one such value, if not to be sure it's the only one).

Then see if you can prove that f(x) > 2 elsewhere. You may want to prove this for several separate cases, based on what the exponents are.

 

[JeffM]

In addition to the ideas provided in previous posts, you might want to consider if there are domains where f(x) is everywhere increasing or everywhere decreasing. A local minimum can occur only at the endpoints of such domains.

 

[Otis]

I like tkhunny's hint. It leads to reasoning that f's minimum point is controlled by its first term only. As horizontal shifts do not affect a function's minimum, I'm thinking that shifting f one unit to the left could make a proof shorter.

?

 

[George Saliaris]

My progress so far : I have proven f is decreasing at (-∞, 1). For x>1 the first term is not a decreasing function..

Edit: So, I took f(x+1) and showed that it is decreasing at (-inf, 0) and increasing at (0, +inf). So f(x+1) has a minimum of 2 for x=0,right?

 

[George Saliaris]

Update No 2 (sorry for the extra post but I could not edit the previous one) : I have a problem showing that f(x+1) is increasing at (0,+inf)..