Question 87821
{{{ f(x)= 1-3x }}} and {{{ g(x)= 1/(x^2+1) }}}


fg(0) says wherevere x is, put zero as its value --> in the g(x) function. So:


{{{ g(x)= 1/(x^2+1) }}} becomes
{{{ g(0)= 1/((0)^2+1) }}}
{{{ g(0)= 1/(0+1) }}}
{{{ g(0)= 1/1 }}}
{{{ g(0)= 1 }}}


Now put this value into f(x):
{{{ f(x)= 1-3x }}} becomes
{{{ f(1)= 1-3(1) }}}
{{{ f(1)= 1-3 }}}
{{{ f(1)= -2 }}}


Now lets work out gf(1). This says to put 1 into f(x) for x. Work out the answer and then put that answer into g(x).


{{{ f(x)= 1-3x }}} becomes
{{{ f(1)= 1-3(1) }}}
{{{ f(1)= 1-3 }}}
{{{ f(1)= -2 }}}
 

then 
{{{ g(-2)= 1/((-2)^2+1) }}} becomes
{{{ g(-2)= 1/(4+1) }}}
{{{ g(-2)= 1/5 }}}


Does this help. Do you understand any better? Read my tutorial on functions - it hopefully explains things in more depth.


cheers
jon