0

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...

minnn
  • 248
  • Haha: A new problem for the CRUDE people and their disciples: Does regurgitating the full textbook solution count as enough "context" to save a question from the axe? – Christian Blatter Nov 12 '18 at 19:20

1 Answers1

0

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

Stockfish
  • 709