Question

F:x→(x+1/x+3), x≠3 and g:x→（6/x-2), x≠2, find f²(4) and g²(1/2)​

Answer 1

Answer:

$f^2(4) = (\frac{25}{49})\\g^2(\frac{1}{2} ) = 16$

Step-by-step explanation:

Here, given:

$f(x) = (\frac{x+1}{x+3} ), g(x) = (\frac{6}{x-2} )$

Now, here  to find the values of $f^2(4), g^2(\frac{1}{2})$

As we know: $f^n(x) = (f(x))^n$

Now, substituting x = 4 in f(x):

$f(4) = \frac{4 +1}{4+3} = \frac{5}{7} \\\implies f^4 = (f(4))^2 = (\frac{5}{7} )^2 = \frac{25}{49} \\\implirs f^{(4)} = ( \frac{25}{49})$

Now, substituting x = 1/2 in g(x):

$g(\frac{1}{2} ) = \frac{6}{\frac{1}{2}-2} = \frac{6}{-\frac{3}{2}} \\\\ g(\frac{1}{2} ) = -6\times (\frac{2}{3} ) = -4\\\implies g^2{(\frac{1}{2}) = ( g(\frac{1}{2} ))^2 = (-4 )^2 =16 \\\\\implies g^2{(\frac{1}{2}) = 16$