Finding the minimum of an exponential function given its domain and maximum

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The domain of function $y=a^{\vert(x-1)\vert+2}$ is $\{x\vert-1\le x\le2\}$, and the maximum value of the function (for the given domain) is $\frac 14$. The problem is the find the minimum value of the function using the given information.

I looked at the solution for this problem and it went like this:

First they take the exponent of the original function as f(x): $f(x)=\vert(x-1)\vert+2$

Using the given domain $\{x\vert-1\le x\le2\}$, it can be deduced that $-2\le x-1\le1$. I understood everything until here, but then it goes on to say that from this it can also be deduced that

$0\le \vert x-1\vert\le2$

I really don't get how this can be deduced...

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1 Answer

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Draw $y \mapsto |y|$. What's the image of $[-2; 1]$?

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