Let f(x)=ax+b, where a and b are real constants, and g(x)=2x-5. Suppose that for all x, it is true that \(g(f(x)) = 3x + 4\). What is a+b?

ant101
Nov 24, 2018

#1**+1 **

I have no idea, but try this:

2(ax+b)-5 is supposed to equal 3x+4, right? From there, it is just algebra.

(don't send hate, I'm 11)

tanmai79
Nov 24, 2018

#5**+3 **

Wait, how about if you equal 2b=4, so b=2. It will be better if you match both sides! Also, 2ax=3x, x will cancel, thus 2a=3, a=3/2. Therefore, 3/2+2=7/2.

ant101
Nov 24, 2018

#8**+3 **

g ( f(x) ) = 2 [ ax + b ] - 5 = 2ax + 2b - 5

So....we have

2ax + 2b - 5 = 3x + 4

This implies that

2ax = 3x ⇒ 2a = 3 ⇒ a = 3/2

And

2b - 5 = 4 add 5 to both sides

2b = 9 divide both sides by 2

b = 9/2

So

a + b = 3/2 + 9/2 = 12 / 2 = 6

CPhill
Nov 24, 2018

#12**+2 **

Another way to approach it is realization!

While we can solve for \(a\) and \(b\) individually, it is simpler to note that \(f(1) = a + b\). Thus, substituting \(1\) into the given equation, we obtain\(g(f(1)) = 3 \cdot 1 + 4 = 7.\)Thus, \(g(f(1)) = 2 \cdot f(1) - 5 = 7 \Longrightarrow f(1) = \boxed{6}.\)

tertre
Nov 24, 2018