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I am given the equation:

$$f(x) = 1+e^{-\cos(x-1)}, [1,2]$$

I graphed this function on wolframalpha and it seems to be continuous on the closed interval $[1,2]$ so that is one thing I am not worried about. I then take the derivative and find it to be:

$$f'(x) = \sin(x−1)e^{−\cos(x−1)}$$

The only critical point I found was $x = 1$, making $\sin(0) = 0$;

I then plug in values $f(1)$ and $f(2)$ to find that the maximum occurs at $f(2)$. However, when I look at the answer key it says there is no absolute max at all. Can someone explain to me why this is so?

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    I edit your question. Is that what you meant? – 2017-01-12
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    Yes thank you :D. – 2017-01-12
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    Every continuous function on a closed interval attains a maximum, so either the answer key is wrong or you're misreading something. – 2017-01-12
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    The textbook is def wrong. Plus, you cannot immediately conclude max. or min. points by $f \prime(x) = 0$ because $1^{st}$ or $2^{nd}$ derivative test was not done. – 2017-01-12
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    Looking at the graph, the function does not have critical points at $0$ or $2$. Even so, over the closed interval in question there are extreme values. Perhaps this is the source of the misunderstanding. – 2017-01-12
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    @AlexMathers To make sure about something, when it's in a closed interval as given [1,2] we still consider f(2) as the absolute max even though outside that interval there is a max greater than it? – 2017-01-12
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    @user3718584 Yes, that's correct. It's the maximum **on the interval $[1,2]$** – 2017-01-12
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    I edited my post to say "absolute max" instead of just "max" at the very bottom in bold, not sure if that changes anyone's answers to this or if it was not clear enough. – 2017-01-12
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    @AlexMathers when you say "maximum" in the reply to my question are you referring to absolute or relative? – 2017-01-12
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    @user3718584 absolute maximum is correct. Every continuous function on a closed interval attains an absolute maximum and minimum. And this function is certainly continuous – 2017-01-12

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$$x\in [1,2]\Rightarrow x-1\in [0,1]\Rightarrow \sin (x-1)\ge 0$$ $$\Rightarrow f'(x)=e^{-\cos (x-1)}\sin (x-1)\ge 0\Rightarrow f\text{ increasing on }[1,2]$$ $$\Rightarrow \begin{cases} f_{\min}(1)=1+e^{-1} \\f_{\max}(2)=1+e^{-\cos 1}. \end{cases}$$