Let f:→R→R,g:R→R and h:R→R be differentiable functions such that f(x)=x3+3x+2,g(f(x))=x and h(g(g(x)))=x for all x ε R. Then
h(0)=16
f(x)=x3+3x+2⇒f′(x)=3x2+3Also f(0)=2,f(1)=6,f(2)=16,f(3)=38,f(6)=236And g(f(x))=x⇒g(2)=0,g(6)=1,g(16)=2,g(38)=3,g(236)=6
(a)g(f(x))=x⇒g′(f(x)).f′(x)=1For g′(2),f(x)=2⇒x=0
∴ Putting x=0, we get g′(f(0))f′(0)=1
⇒g′(2)=13
(b) h(g(g(x)))=x⇒h′(g(g(x))).g′(g(x)).g′(x)=1For h′(1) we need g (g(x)) = 1
∴g(x)=6⇒x=236∴ Putting x = 236, we get h′[g(g(236))]=1g′(g(236)).g′(236)⇒h′(g(6))=1g′(6).g′(236)⇒h′(1)=1g′(f(1)).g′(f(6))=f′(1).f′(6)=6×111=666
(c) h[g(g(x))]=xFor h(0),g(g(x))=0⇒g(x)=2⇒x=16
∴ Putting x = 16, we get
h(g(g(16)))=16⇒h(0)=16
(d)h[g(g(x))]=xFor h(g(3)),we need g(x)=3⇒x=38
∴ Putting x = 38, we get
h[g(g(38))]=38⇒h(g(3))=38