If f- - 6- 6- - R is defined by f x - x3 - 3- - x - R- then fofof - 1 - fofof - 1 - fofof 0 - fofof 0 -

f:[ 6 ,6] → R f( x ) = x^3 3 now, ( fofofo)( 0 ) = f( f( f (x) )) = f( f( x^3 3)) = f( ( x^2 3)^2 3) = [ ( x^2 3)^2 3]^2 3 ...

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If $f : [- 6, 6] \to R$ is defined by $f (x) = x^{3} - 3, \forall x \in R,$ then $(f o f o f) (- 1) + (f o f o f) (- 1) + (f o f o f) (0) + (f o f o f) (0) =$

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f : [- 6, 6] \to R

f (x) = x^{2} - 3

x \in R

(f o f o f) (- 1) + (f o f o f) (0) + (f o f o f) (1) =

f : R \to R

f (x) = \frac{x}{\sqrt{1 + x^{2}}}

(f o f o f (x))

f (x) = \frac{x}{\sqrt{1 + x^{2}}}

f : [- 6, 6] \to R

f (x) = x^{2} - 3, \forall x \in R

(f \circ f \circ f) (- 1) + (f \circ f \circ f) (0) + (f \circ f \circ f) (1) =

f

R

R

f (x) = x^{3} - 1

f^{- 1} {- 2, 0, 7} =

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