The functions f(x) and g(x) are given by f(x)=2x g(x)=3+2x (a) Calculate the value of gf(4) (b) Solve the equation fg(x)=14?

(a) Calculate the value of gf(4)

g(f(x)) = g(2x)

= 3 + 2(2x)

= 3 + 4x


(b) Solve the equation fg(x)=14

f(g(x) = f( 3 + 2x)

= 2 (3 + 2x)

= 6 + 4x

6 + 4x = 14

4x = 14 - 6

= 8

x = 8/4

= 2

Why, oh why, do they make kids do these problems?
In my nearly half century of engineering experience, I have never had a need for this sort of operation (nor have I ever had to change bases in logarithms, nor have I needed to perform a trigonometric substitution when solving an integral)

I assume that fg means f(g(x))

gf(4) = g(f(4)) = g(2*4) = g(8) = 3 + 2*8 = 19


f(g(x)) = 14
2* g(x) = 14
g(x) = 7
3 + 2x = 7
2x = 4
x = 2

g(f(x)) => 3 + 2(2x) = 3 + 4x

so, g(f(4)) => 3 + 4(4) = 19

f(g(x)) => 2(3 + 2x) = 6 + 4x

Hence, 6 + 4x = 14

=> 4x = 8

so, x = 2

:)>

Well,

I think brackets are missing ...

f(x)=2x
and
g(x)=3+2x
so
g(f(4)) = g(2*4) = g(8) = 3 + 2*8 = 19
f(g(x)) = f(3+2x) = 2*(3+2x) = 6 + 4x
so we want to solve
6 + 4x = 14
4x + 6 = 14
4x + 6 - 6 = 14 - 6
4x + 0 = 8
1/4 * 4 * x = 8/4
1 * x = 2
x = 2 <--- answer

qed

hope it' ll help !!

PS: Welcome to the forum !!
if you want good answers, don't forget to give BAs too !! ;-)