SAT Math Question? (from May 2012 SAT)

RandomHSer
Tue, 10 Jul 2012 14:20:12 +0000

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