Question

Given that f(x) = 3x + 1 and g(x) = the quantity of 4x plus 2 divided by 3 solve for g(f(0)).

Answer 1

The best way to think about this problem is what are you plugging int to what?

g(f(0)) means, you plug 0 into f(x).

Whatever number you get from that, you plug into g(x)!

Give it a try and let me know what you get

Answer 2

so right now i got g(f(0)=\[\frac{ 4(3x+1)+2 }{ 3 }\]

Answer 3

First let's find F(0).

Plug in 0 for x in F(x)

F(x) = 3x +1
F(0) = 3(0) + 1
F(0) = 0 +1
F(0) = 1

So now, you plug in 1 into g(x)

G(1) = ?????

Answer 4

G(x)=\[\frac{ 4x+2 }{ 3 }\]
G(1)=\[\frac{ 4(1)+2 }{3}\]
G(1)=\[\frac{ 6 }{ 3 }\]
G(1)=\[\frac{ 2 }{ 1 }\] = 2
G(1)=2

Answer 5

Exactly!

So g(f(0)) = 2

Remember the procedure and you will do fine!

Answer 6

g(f(0)) = g(1) = 2

Answer 7

can you help me with this one Given f(x) = the quantity of 2x+4/
5 solve for f-1(3).?

Answer 8

I don't understand the very last part of your problem

"f-1(3)"

Can you rephrase that?

Answer 9

idk that what the paper says i think it means the inverse