MHB Optimisation Problem (Global extreme points)

[Foosey96]

Hi there everyone, wonder if anyone can help as I'm a bit confused.
Ive been asked to find the global extreme point of f(x)=e^(x-1) - x.
I have checked my answer against the solution and am correct and my working is as follows:
f'(x) = e^(x-1) - 1 = 0. Therefore (x-1)=ln1 (which = 0) therefore x = 1.
f''(x) = e(x-1). sub in x=1 and f''(x) = 1 which is > 0 hence the extreme point is a minimum.
The solution however then goes on to conclude that x*=1 minimizes f(x)
This is what I am confused by as I thought minimum points by their nature were the y-values not the x values and hence x* should actually be e^(1-1) - 1 = 0 and so x*=0 minimizes f(x)??
Thanks for any help you can give that final conclusion has just thrown me off. Thanks everyone!

 

[Prove It]

Hi there everyone, wonder if anyone can help as I'm a bit confused.
Ive been asked to find the global extreme point of f(x)=e^(x-1) - x.
I have checked my answer against the solution and am correct and my working is as follows:
f'(x) = e^(x-1) - 1 = 0. Therefore (x-1)=ln1 (which = 0) therefore x = 1.
f''(x) = e(x-1). sub in x=1 and f''(x) = 1 which is > 0 hence the extreme point is a minimum.
The solution however then goes on to conclude that x*=1 minimizes f(x)
This is what I am confused by as I thought minimum points by their nature were the y-values not the x values and hence x* should actually be e^(1-1) - 1 = 0 and so x*=0 minimizes f(x)??
Thanks for any help you can give that final conclusion has just thrown me off. Thanks everyone!

What they said is correct. To say that x = 1 minimises f(x) is to say that when x = 1 IN the function, then the function value will be at its minimum.

Here the minimum value is f(1) = e^(1-1) - 1 = -1.