maths question - functions

here's my working out:
1) Find the value of C
fg(x) = f(cx+4)
= 2(2c+4) + (c+4)
=2cx + 12 + c

gf(x) = g(2x+c)
= (2x+c)+4
= 2x + c +4

fg(x) = gf(x)
C = 4

then D should be 12 right?

fg(x) = f(cx+4)
= 2(2c+4) + (c+4)
=2cx + 12 + c

The line in red is incorrect. Can you check it again?

2(cx + 4) + (c + 4) i think, yea just a typing error

p.s how do i change my profile image?

In general, $fg(x) \neq gf(x)$ so you can't use that as a rule.
You're right that c = 4, though.

p.s how do i change my profile image?

??? what's the right way then?

$fg(x) = f(cx + 4) = 2(cx+4) + c = 8x + d$

When you stack functions, g(x) becomes the unknown variable in f(x)

The (c+4) shouldn’t be there, it should still just be c. When doing fg(x), you’re substituting g(x) in f(x) ONLY where there is x.

'0_o I'm so confused right now!!!!!!!!!!!e got 2 I've got 2 question:
1) i though that fg(x) = f(cx + 4) = 2(cx + 4) + (c+4) - Am I wrong?
2) Why did you only used 2cx instead of 2cx + 8 = 8x?

The question says that c and d are constants. They can't vary.

Okay, let's start at the beginning - f(x) means a function of x, where x is the variable in the function. You take variable x, you do a bunch of things to it, you've got a function of x.
So we've got two functions, f(x) and g(x). They do different things to the same variable.

Whatever you put into the brackets - that's called the argument - is the variable that's affected. So $f(x) = 2x + c, f(t) = 2t + c, f(a) = 2a + c.$
Note, that 'c' is the same no matter what you do - because the question said that c was a constant. Only one part of the function has the variable.
So, what happens if you put a function in as the argument? Well, that's $f(g(x))$. And you know that $g(x) = cx + 4$.
So you can write that as $f(cx + 4)$. And when you set the variable to $cx + 4$, you get $2(cx + 4) + c$

The question says that c and d are constants. They can't vary.

Okay, let's start at the beginning - f(x) means a function of x, where x is the variable in the function. You take variable x, you do a bunch of things to it, you've got a function of x.
So we've got two functions, f(x) and g(x). They do different things to the same variable.

Whatever you put into the brackets - that's called the argument - is the variable that's affected. So $f(x) = 2x + c, f(t) = 2t + c, f(a) = 2a + c.$
Note, that 'c' is the same no matter what you do - because the question said that c was a constant. Only one part of the function has the variable.
So, what happens if you put a function in as the argument? Well, that's $f(g(x))$. And you know that $g(x) = cx + 4$.
So you can write that as $f(cx + 4)$. And when you set the variable to $cx + 4$, you get $2(cx + 4) + c$

ohhh now i understand! one last question: is the answer to this question 12?

Bruh that’s complicating just it 😂😂

If question 12 is what you posted (can't see the question number in the screenshot) then that's pretty much most of the work for it.



Ay don't knock it, it worked

no i mean is d = 12?

Yeah