Question

f(x)=x^2 when x>=0 and f(x)=-x^2 when x <0. find 2f(x)>=f(4-x^2)

Answer 1

Ques solution x2 N 7,0 Given Yolution flm) -12 nco then. N TO 2012 2 fon) -2 n2 i aco graph of 2 fm) will be an follows 1 y-axis K 1 N-axis

Now deduce from f 14-32) given function for we can an, 4-112 to -☺ f (4-2) = (4-02)2 -(4-22) 2 ; 4-12 co Now, 4.227,0 -) 212<4 =) 171) < 2 4.712 Lo -) 12 > 4 then 1 lil 22 = 1072 equation reduces to f (4-72) = (4-21 22 B -14-232 101 > 2 then graph of f(4-72) will be as follows, y-axis 10,16) L 0 2 N-axis

How to find intersection point of 2 fois and 2 (4-22) 2 (4-2232 -) ㅕ tun?) 222 = (4-72) for OLN<2 and -2012 for n<-2 2012 = 2012 = 16 +314-872 24-1002 H16 50 nu_on² - 2n2 +16=0 =) (212-8) (a12-2) I 2 t2d2 Using domain restrictions get inter section points (2,0) (-2,2,0). Therefore combined graph of 2f101) and f (4-02) will be we as as follows, Yraxis (12,4) + Hoaxin (-2,72716) Therefore from figure it is clear af (81) > f 14-42) for af (-20, -212] U (3,0)