MHB Optimisation Problem (Global extreme points)

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The discussion revolves around finding the global extreme point of the function f(x) = e^(x-1) - x. The derivative f'(x) is set to zero, leading to the conclusion that x = 1 is a critical point, which is confirmed as a minimum by the second derivative test. The confusion arises from interpreting the term "minimizes f(x)," where it is clarified that this refers to the x-value at which the minimum occurs, not the minimum value itself. The minimum value of the function at this point is f(1) = -1. Thus, x = 1 is the correct point that minimizes the function.
Foosey96
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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!
 
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Foosey96 said:
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.
 
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