+0  
 
+2
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avatar+742 

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
avatar+135 
+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
 #2
avatar+742 
+3

That's what I did! How do you proceed from there, though?

ant101  Nov 24, 2018
 #3
avatar+135 
+1

I don't know, it's probably some complicated stuff that needs some sort of manipulation. I'm only a 6th grader doing 7th grade level math, and one who will be doing 9th grade level math next semester, okay?

tanmai79  Nov 24, 2018
 #4
avatar+135 
+2

Some of the moderators could probably solve this.

tanmai79  Nov 24, 2018
 #5
avatar+742 
+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
 #6
avatar+135 
+2

are you allowed to just do that, though? Just put in a random value?

tanmai79  Nov 24, 2018
 #7
avatar+742 
+3

Hmm, if you match both equations... Let's see what CPhill says.

ant101  Nov 24, 2018
 #8
avatar+92856 
+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

 

 

cool cool cool

CPhill  Nov 24, 2018
 #9
avatar+135 
+2

So our equation was right, we just couldn't figure out how to solve it.

tanmai79  Nov 24, 2018
 #10
avatar+742 
+3

Oh, thank you, CPhill! I totally forgot about the -5!

ant101  Nov 24, 2018
 #11
avatar+135 
+1

I don't know much about functions, but this makes sense.

tanmai79  Nov 24, 2018
 #12
avatar+3478 
+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
 #13
avatar+92856 
+1

Thanks, tertre....!!!

 

 

cool cool cool

CPhill  Nov 24, 2018

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