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# Functions Help!

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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?

Nov 24, 2018

#1
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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)

Nov 24, 2018
#2
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That's what I did! How do you proceed from there, though?

ant101  Nov 24, 2018
#3
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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
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Some of the moderators could probably solve this.

tanmai79  Nov 24, 2018
#5
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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.

Nov 24, 2018
#6
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are you allowed to just do that, though? Just put in a random value?

tanmai79  Nov 24, 2018
#7
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Hmm, if you match both equations... Let's see what CPhill says.

ant101  Nov 24, 2018
#8
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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

Nov 24, 2018
#9
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So our equation was right, we just couldn't figure out how to solve it.

tanmai79  Nov 24, 2018
#10
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Oh, thank you, CPhill! I totally forgot about the -5!

ant101  Nov 24, 2018
#11
+142
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I don't know much about functions, but this makes sense.

tanmai79  Nov 24, 2018
#12
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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, $$g(f(1)) = 2 \cdot f(1) - 5 = 7 \Longrightarrow f(1) = \boxed{6}.$$

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Nov 24, 2018
#13
+107405
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Thanks, tertre....!!!

CPhill  Nov 24, 2018