## How Do You Evaluate a Function? II

$\fn_jvn&space;{\color{Blue}&space;f(x)=x^2-2x+4}$This post about how to evaluate a function is part of a series of posts to help you prepare for the Advanced Algebra and Functions part of the Accuplacer test.

## Question

For the function

$\fn_jvn&space;{\color{Blue}&space;f(x)=x^2-2x+4}$

what is the value of $\inline&space;\fn_jvn&space;f(x-1)$?

A.  $\inline&space;\fn_jvn&space;x^2-2x+3$

B. $\inline&space;\fn_jvn&space;(x-1)^2-2(x-1)+4$$\inline&space;\fn_jvn&space;(x-1)(x^2-2x+3)$

C. $\inline&space;\fn_jvn&space;(x-1)(x^2-2x+3)$

D.  $\fn_jvn&space;x^2+3x+4$

## Solution

f(x-1) is what you get when you replace x with x-1 in $\inline \fn_jvn x^2-2x+4$. That is called “evaluating the function for x-1.

### Sidebar: Function Notation

In algebra, usually when you write two things side by side with one of them in parentheses, that means you multiply those two things together:

$\fn_jvn&space;{\color{Blue}&space;a(b)=a&space;\cdot&space;b}$

So no one can blame you for thinking f(x) means f multiplied by x. But math notation, like other kinds of language, has exceptions. f(x) means function f, carried out on x. And f(something else) means function f carried out on something else. For example, say function g is defined as $\inline \fn_jvn g(x)=x^2$. That means $\inline \fn_jvn g(1)=1^2$; $\inline \fn_jvn g(-2)=\left ( -2 \right )^2$; and $\fn_jvn g(*)$ means $\inline \fn_jvn *^2$ (whatever * means).

It gets interesting when the parentheses in the function definition contain a function. For example, if  $\inline \fn_jvn g(x)=x^2$, what is $\inline \fn_jvn g(x+2)$? Well, we said that if you want to find $\fn_jvn g(*)$, you insert * into the function definition. To find $\inline \fn_jvn g(x+2)$, insert $\inline \fn_jvn x+2$ into the function definition: if $\inline \fn_jvn g(x)=x^2$, then $\inline \fn_jvn g(x+2)=(x+2)^2$.

So to find f(x-1) in the expression

$\fn_jvn&space;{\color{Blue}&space;f(x)=x^2-2x+4}$

insert x-1 wherever you see x. You should get:

$\fn_jvn&space;{\color{Blue}&space;(x-1)^2-2(x-1)+4}$

Usually an answer like this would be expressed as $\inline \fn_jvn x^2-4x+7$, but this question is multiple choice so you go with what you’ve got.