Question 1:
The function f and g is defined by
f:x↦2x−3g:x↦2x;x≠0
Find the expression for each of the following functions
(a) ff,
(b) gf,
(c) f-1 ,
Calculate the value of x such that ff(x) = gf(x).
Solution:
(a)
ff(x)=f[f(x)] =f(2x−3) =2(2x−3)−3 =4x−9ff:x↦4x−9
(b)
gf(x)=g[f(x)] =g(2x−3) =22x−3gf:x↦22x−3
(c)
Let f−1(x)=y,thus f(y)=x 2y−3=x y=x+32∴ f−1(x)=x+32f−1:x↦x+32When ff(x)=gf(x),4x−9=22x−3(4x−9)(2x−3)=28x2−30x+27=28x2−30x+25=0(4x−5)(2x−5)=04x−5=0 or 2x−5=0x=54 or x=52
The function f and g is defined by
f:x↦2x−3g:x↦2x;x≠0
Find the expression for each of the following functions
(a) ff,
(b) gf,
(c) f-1 ,
Calculate the value of x such that ff(x) = gf(x).
Solution:
(a)
ff(x)=f[f(x)] =f(2x−3) =2(2x−3)−3 =4x−9ff:x↦4x−9
(b)
gf(x)=g[f(x)] =g(2x−3) =22x−3gf:x↦22x−3
(c)
Let f−1(x)=y,thus f(y)=x 2y−3=x y=x+32∴ f−1(x)=x+32f−1:x↦x+32When ff(x)=gf(x),4x−9=22x−3(4x−9)(2x−3)=28x2−30x+27=28x2−30x+25=0(4x−5)(2x−5)=04x−5=0 or 2x−5=0x=54 or x=52