# Composite function

• Oct 11th 2007, 11:04 PM
DivideBy0
Composite function
If $\displaystyle f(x)=\frac{1}{x-1}$ for $\displaystyle x>2$ and $\displaystyle f(x)=x+1$ for $\displaystyle 2 \geq x$, find $\displaystyle f(a-2)$ in terms of a.

Normally I would simply substitute a in, but I don't know if that is the right way to go about things?

I think I should have considered pre-calc a little more carefully before going into calc :)
• Oct 11th 2007, 11:21 PM
earboth
Quote:

Originally Posted by DivideBy0
If $\displaystyle f(x)=\frac{1}{x-1}$ for $\displaystyle x>2$ and $\displaystyle f(x)=x+1$ for $\displaystyle 2 \geq x$, find $\displaystyle f(a-2)$ in terms of a.

Normally I would simply substitute a in, but I don't know if that is the right way to go about things?...

Hello,

of course your plan is OK - but you have to plug in the term (a-2) into the conditions too:

$\displaystyle f(x)=\left \{\begin{array}{lr}\frac{1}{x-1} & x>2 \\x+1 & x\leq 2\end{array}\right.$

Now you get:

$\displaystyle f(a-2)=\left \{\begin{array}{lr}\frac{1}{a-2-1} & a-2>2 \\a-2+1 & a-2\leq 2\end{array}\right.$ After simplifying a little bit you have:

$\displaystyle f(a-2)=\left \{\begin{array}{lr}\frac{1}{a-3} & a>4 \\a-1 & a\leq 4 \end{array}\right.$