SAT Math Question? (from May 2012 SAT)

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<p>Hi, </p>

<p>This is my first time posting on this forum. I hope I'm posting in the right place! I sometimes come here to find answer explanations to SAT problems that I don't understand~ (so helpful!) I'm currently reviewing some of the math problems I got wrong on the May 2012 SAT's, and I'm stumped on one question. Help would be highly appreciated :)</p>

<h2>Can someone please EXPLAIN the answer to this question? VV</h2>

<p>"The function f is defined by f(x) = 2x^2 - 5. What are all possible values of f(x) where -2<x<2?"</p>

<p>(a) -5 ≤ f(x) < 0
(b) -5 ≤ f(x) < 3
(c) 0 ≤ f(x) < 3
(d) 0 ≤ f(x) < 8
(e) 2 ≤ f(x) < 8</p>

<p>The answer is B. (I got A.)</p>

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<p>f(x)=2x^2-5. This has a minimum point at (0,-5) since 2x^2 is never less than zero, and only equal to zero when x=0. The function is increasing for x>0 and plugging in the value of x with the largest absolute value will give the maximum (sign doesn’t matter since f is an even function). Plugging in x=+/-2 we get f(+/-2)=8-5=3. Since f is continuous as x approaches +/- 2, f(x) approaches this value. If f has a min at -5 and approaches 3 as a max then -5<=f(x)<3</p>