If 0 < x ≤ 2, then which one of the following is the maximum value of (x – 1)^2 + x ? Choose one:  A) –2  B) –1  C) 0  D) 2  E) 3
If you cannot spot that the maximum value of (x – 1)^2 + x for 0 &lt; x ≤ 2 is when x=2 --&gt; (x – 1)^2 + x = 2 you can do the following:   Since 0 &lt; x ≤ 2, then (x – 1)^2 + x is positive --&gt; discard A, B and C.   For the same reason: (x – 1)^2 is a non-negative fraction less than 1.   Thus (x – 1)^2 + x = (non-negative fraction less than 1) + (0 &lt; x ≤ 2) &lt; 3. Discard E.   Only D is left.   Answer: 2.
<result>D</result>
If 0 < x ≤ 2, then which one of the following is the maximum value of (x – 1)^2 + x ?