For each value of y greater than 23, the function f(x) is

Could you please elaborate more on your question; I have not understood it.

I am confused as to how we got b=4. If I plug in 4 for "b", the answer is 7. However, if I plug in a=4, and b=6, the answer becomes f(4)=4.
Where did I go wrong?

You may be right, I did not like the question so much. Did you mean to say "a = 8 and NOT 4"? b = 6 seems to be ok. kindly show your calculations.

Yes! Correct, I meant to type a=8, b=6. Thank you so much!

I am confused as to how we got b=4

­A few things I noted while reading the question:


1. 2√3 is slightly less than 4.
2√4 = 4. So, 2√3 would be slightly less.

2. x will be greater than y.
The given equation can be simplified to x = y + y/12
Since y is positive, x will be greater than y.

3.
The function is written in the form of 'y = <some function of x>', but the equation is then in the form of 'x = 'some expression in terms of y'

I needed a moment to ensure I processed it correctly.

The given equation in terms of x and y is essentially an elaboration of the relation between x and y. f(x) = y tells us that there is a relation between x and y. What that relation is we get from the equation.

4. 'a' maps to 'x' and 'b' maps to 'y'
I realized that I could mix up the two constants and end up getting the answer wrong. So, I spent extra time to ensure I understood what a and b represent.

f(x) = y
maps with
f(a) = b
So, 'a' maps to 'x' and 'b' maps to 'y'.

Since I had already understood that x > y, 
I realize a > b.


When I look at the table

1. I can reject 1 and 2 in b's column. Since y > 2√3
2. I can reject 8 for b. Since a > b, so b can't have the highest value.
3. I have 2 options left for b. 4 and 6. I can plug them in.

i. b = 4
x = y + 12/y

if b = 4,
a = 4 + 12/4 
a = 4 + 3 = 7.

7 isn't an option for a.

Reject 4 for b.

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If I were running short on time at this point, I could directly mark b = 6 and a = 8 at this point. 

Why a = 8?

Because a > b. So, if b = 6 (which is the only feasible value left for b), a has to be greater than 6. That leaves only one choice for a.

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ii. b = 6
x = y + 12/y

if b = 6,
a = 6 + 12/6
a = 6 + 2
a = 8.


a = 8, b = 6.
 

 

­ \(x=\frac{y^2+12}{y}\) or  \(a=\frac{b^2+12}{b}=b+\frac{12}{b}\)

Surely a>b and \(b>2\sqrt{3}>2\).
Also, both a and b are even. 
\(a=b+\frac{12}{b}\) means 12/b is even as even = even + 12/b

12/4 =3...Discard
12/6 = 2..Yes

so \(8=6+(\frac{12}{6})\)