Composite function

**DivideBy0** · October 11th 2007, 11:04 PM

If $f(x)=\frac{1}{x-1}$ for $x>2$ and $f(x)=x+1$ for $2 \geq x$ , find $f(a-2)$ 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

**earboth** · October 11th 2007, 11:21 PM

Originally Posted by DivideBy0

If $f(x)=\frac{1}{x-1}$ for $x>2$ and $f(x)=x+1$ for $2 \geq x$ , find $f(a-2)$ 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:

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

Now you get:

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

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

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