Question

if f:[-6,6] belongs to R is defined by f(x)= x^2-3 where x belongs to R,then(fofof)(-1)+(fofof)(0)+(fofof)(1)=​

Answer 1

SOLUTION

GIVEN

f : [ - 6, 6 ] → R is defined by

$\sf{f(x) = 2 {x}^{2} - 3}$

TO DETERMINE

$\sf{(f \circ f \circ f)( - 1) + (f \circ f \circ f)( 0) + (f \circ f \circ f)( 1) }$

EVALUATION

Here it is given that

$\sf{f(x) = 2 {x}^{2} - 3}$

Now

$\sf{(f \circ f \circ f)( - 1) }$

$\sf{ = f ( f ( f( - 1))) }$

$\sf{ = f ( f ( - 1))}$

$\sf{ = f ( - 1)}$

$= - 1$

Again

$\sf{ f( f ( f ( 0)))}$

$\sf{ = f( f ( - 3))}$

$\sf{ = f( 15)}$

$= 447$

Again

$\sf{(f \circ f \circ f)( 1) }$

$\sf{ = f ( f ( f( 1))) }$

$\sf{ = f ( f ( - 1)) }$

$\sf{ = f ( - 1) }$

$= - 1$

Hence

$\sf{(f \circ f \circ f)( - 1) + (f \circ f \circ f)( 0) + (f \circ f \circ f)( 1) }$

$= - 1 + 447 - 1$

$= 445$

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