consider function f(x)

Question

Consider the function f(x) = 3/2-x , x ≠ 2 and g(x)= 2- √2x-3 , x ≥ k

k is a constant. Find:

1) the value of k

2) (f ^-1 ° g)(2)

Answer 1

1) the value of k

f(x) = 3/(2-x) x≠ 2

g(x) = 2 - √(2x-3) so x≥k, so 2x-3 ≥0, 2x≥3, x≥3/2 k=3/2

2) (f ^-1 ° g)(2)

f^-1 → y = 3/(2-x), switching y and x yields f^-1 = (2x-3)/x so x≠ 0

and (f ^-1 ° g)(2) = 0

Answer 2

Consider the function f(x) = 3/2-x , x ≠ 2 and g(x)= 2- √2x-3 , x ≥ k

k is a constant. Find:

1) the value of k

2) (f ^-1 ° g)(2)

SOLUTION

κ ≥ 3/2

f(x) = 3/(x-2) ⇒ f^-1( x ) = 2 +(3/x)

g( x) = 2 - √(2x -3) ⇒ g^-1 (x) = ( x² -4x +7 ) / 2

( f^-1 ο g ) (2) = f^-1(g(2)) = f^-1( 1 ) = 5