Not Understanding a specific substitution rule I was given the question,
If $f(3x+5) = x^2-1$, what is $f(2)$? I am trying to understand the reasoning why $3x+5$ is set equal to $2$.
 A: This problem is something of a "trick".
If you knew a formula for $f(x)$, you could just substitute $2$ for $x$ to get the answer directly. However, they don't make it that easy -- you only know a formula for $f(3x+5)$. That means you have to figure out what to use for $x$ in order to have the expression that's plugged into $f$ evaluate to $2$ (so that you are still computing $f(2)$ as desired):
$$f(\;\overbrace{3x+5}^{\textrm{what makes this }2?}\;)$$
So you need to find $x$ such that $3x+5=2$, i.e., solve $3x+5=2$ for $x$. Doing this gives $x=-1$. Then substitute that value ($-1$) for $x$ to get
$$f(3(-1)+5 ) = (-1)^2-1$$
$$f(2) = 0$$
A: To find $f(2)$, we want to consider $3x+5=2$, or $x=-1$.
Then we have $f(3(-1)+5)=(-1)^2-1=0$.
A: Let's say $u=3x+5$. Then, $x^2-1=(\frac{u-5}{3})^2-1$. $\Rightarrow$ $f(u)=(\frac{u-5}{3})^2-1$ $\Rightarrow$  $f(x)=(\frac{x-5}{3})^2-1$. Now we can easily see that $f(2)=0$ . However, instead of doing all these complicated stuff, we could have made the following deduction: $3x+5=2$ $\Rightarrow$ $x=-1$ $\Rightarrow$ $x^2-1=0$. To summarize, $3x+5=2 \Rightarrow x^2-1=0$ $\:\:\:\equiv\:\:\:$ $f(2)=0.$
SIMILAR EXAMPLE
$f(x-1)=x^2\Rightarrow f(0)=?$
Say $u=x-1 \Rightarrow (u+1)^2=x^2 \Rightarrow f(u)=(u+1)^2 \Rightarrow f(x)=(x+1)^2 \Rightarrow f(0)=1.$
OR
Set $$x-1=0 \Rightarrow x=1 \Rightarrow x^2=1$$
$$\Downarrow$$
$$x-1=0 \: and \: x^2=1 \:\:\equiv\:\: f(0)=1$$
A: The problem here is with the mindset that looks at $3x+5$ and thinks "the result of 3 times THE number plus 5". Another way to look at this problem is "I'm thinking of a number ($3x+5$). What would $x^2-1$ be?". If we think of $3x+5$ as THE number, then we can look at $x^2-1$ with new eyes.
$$x^2-1 = \bigg(\frac{(3x+5)-5}{3}\bigg)^2-1$$
In this case, we try to find out what $x$ would be in terms of THE number $3x+5$.
This mental trap develops because teachers gloss over the Closure Principle in middle and secondary schools. While everybody understands that if $x$ is a number then $3x$ is also a number, we tend to see the latter expression as a different animal than the former when solving problems. We see $x$ as a number but $3x$ isn't; it's an expression that I have to evaluate. In reality, both are numbers, and both are expressions. I can interpret $3x$ as an expression to evaluate, just as easily as I can $\frac{3x}{3}$ (i.e. $x$). Similarly I can ask myself, if $3x$ equals $5$, what does $3x+1$ equal? In this case there is absolutely no need find $x$ when I already know the value of $3x$. Why? Because $3x$ is a number.
A: Because it will alow you to put which $x$ needs to be placed on $3x+5$ in order to get $f(2)$. So having $3x+5=2$, then $x=-1$, so this means we need to put $x=-1$ on $3x+5$ to get $f(2)$, that is $f(3(-1)+5)=f(2)=(-1)^2-1=0$, as Elliot G shown.
A: In essence, it says to find $f(2)$. We have the function value defined at $f(3x+5)$. We try to solve the following equation $$f(3x+5) = f(2)$$ $$3x+5 = 2$$ $$3x = -3$$ $$x = -1.$$
So thus if we substitute in $x=-1,$ we get $$\begin{align}f[3(-1) + 5] & = f(2) \\&= (-1)^2 - 1 \\&= 1 - 1 \\&= 0.\end{align}$$
So $f(2) = 0.$
