Given f(x) = √x - 2, find: 3f(4(x-1)) + 5
I cannot get started.
Think of f as "doing something" to what's in the bracket.
In this case f means "take the square root and subtract 2".
So look at what you have:
$\displaystyle 3f(4(x-1)) + 5$
In the brackets defining the f there's $\displaystyle 4(x-1)$.
So do the square root of all that and subtract 2. Then when you done that, multiply the whole thing by 3 and then add 5.
i still cant get it man
can u show me a step by step?
this is how far i get
Given f(x) = √x - 2, find: 3f(4(x-1)) + 5
f(x) = √(4x-4)
Basically I dumbed down the brackets of (4(x-1)) to (4x-4) and put that in for x, but I don't know what to do from there....