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(A) f(x) has a relative minimum at x = 0 f(x) is concave downwards for all x has a point of inflection at (0, f (0))(b) If g(a) 6= 0, then f=gis continuous at x= a Theorem 3 If fis continuous at a, and if gis continuous at f(a), then f g is continuous at a Intermediate Value Theorem (IVT) Suppose fis continuous on a;b If kis any number between f(a) and f(b), then there exists a numberWhere G0 is an operator which does not depend on x Checking the special case x= 0, we find G= 1 and F=0 Therefore G0 =1 Eliminating Fand Gin favor of nd B, we find exA exB = exAxB 1 2 x2 A,B Setting x=1, this becomes Secure Media Collegeboard Org ƒiƒ`ƒ…ƒ‰ƒ‹ ƒ~ƒfƒBƒAƒ€ ƒXƒgƒŒ[ƒg