f.4 functions

6.It is given that f(x) = 3x - 2.
(a) Find the values of f(2) and f(4).
f(x) = 3x - 2
f(2)=3(2)-2=4
f(4)=3(4)-2=10
(b)
[f(2)]^2
= [(3*2 - 2)]^2
=16
f(2^2)
=f(4)
=10(as calculated above)
so [f(2)]^2 ≠ f(2^2)
20a. It is given that f(x) = (x + 3)(x + 2) - k^2 and f (k) = 2k.
f(x) = (x + 3)(x + 2) - k^2
by substituting k into the equation
f(k) = (k + 3)(k+ 2) - k^2
f(k) = k ^2+ 5k+ 6 - k^2
f(k) = 5k+ 6
and as f(k) also equal to 2k,so
2k=5k+6
-3k=6
k=-2
b/Hence, solve for x if 2x - f(x) = 0.
2x-f(x)=0
2x- [(x + 3)(x + 2) - k^2 ]=0
2x- [(x^2+5x+6 - k^2 ]=0
2x-x^2-5x-6+(-2)^2=0
-x^2-3x-2=0
x^2+3x+2=0
(x+2)(x+1)=0
x=-2 or x=-1

6a)f(2) = 3(2) - 2
= 6- 2
= 4
f(4) = 3(4) -2
= 12 -2
= 10