3.03 Review: Evaluating functions

Lesson

Practice

Lesson

In the lesson Functions and Relations, we were also introduced to the concept of functions, where each input yielded a unique output.

When we are writing in function notation, instead of writing "$y=$y=", we write "$f(x)=$f(x)=". This gives us a bit more flexibility when we're working with equations or graphing as we don't have to keep track of so many $y$ys! Instead, using function notation, we can write $f(x)=$f(x)=, $g(x)=$g(x)=, $h(x)=$h(x)= and so on. These are all different expressions that involve only $x$x as the variable.

We can also evaluate "$f(x)$f(x)" by substituting values into the equations just like we would if the question was in the form "$y=$y=".

Worked example

Question 1

If $f(x)=2x+1$f(x)=2x+1 , find $f(5)$f(5).

Think: This means we need to substitute $5$5 in for $x$x in the $f(x)$f(x) equation.

Do:

$f(5)$`f`(5)	$=$=	$2\times5+1$2×5+1
	$=$=	$11$11

Reflect: Check the reasonableness of your calculation. Does it make sense?

Practice questions

Question 2

If $f\left(x\right)=4x+4$f(x)=4x+4,

find $f\left(2\right)$f(2).
find $f\left(-5\right)$f(−5).

Question 3

Consider the function $p\left(x\right)=x^2+8$p(x)=x2+8.

Evaluate $p\left(2\right)$p(2).
Form an expression for $p\left(m\right)$p(m).

Question 4

Use the graph of the function $f\left(x\right)$f(x) to find each of the following values.

Loading Graph...

A line with a positive slope is plotted in a cartesian coordinate plane. The line passes through points $\left(-2,-1\right)$(−2,−1), $\left(-1,0\right)$(−1,0), $\left(0,1\right)$(0,1), $\left(1,2\right)$(1,2), and $\left(2,3\right)$(2,3), as represented by the dots. The line crosses the x-axis at point $\left(-1,0\right)$(−1,0), and the y-axis at point $\left(0,1\right)$(0,1).

$f\left(0\right)$f(0)
$f\left(-2\right)$f(−2)
Find the value of $x$x such that $f\left(x\right)=2$f(x)=2