EVALUATION OF FUNCTION

Before evaluating a function, let review some important concepts.

A function is a relation in which each valid input value leads to exactly one output value.

One way of writing function is function notation. A function notation shows the functional relationship between the input value and output value. 

The most common function notation is:

$f(x)$

Note: This is read of "f of x", not f times x.

Evaluation of function means determining the output value for a given input value. Stated differently, function evaluation means we should obtain output value for a given input value.

Example 1

If $f(x)=x^2$, find

i) $f(1)$

ii) $f(2)$

iii) $f(0)$

Solution:

i) $f(1)$ means we should insert $1$ for x

$f(1)=(1)^2$

$f(1)=1$

ii) $f(2)$ means we should insert $2$ for x

$f(2)=(2)^2$

$f(2)=4$

iii) $f(0)$ means we should insert $0$ for x

$f(0)=(0)^2$

$f(0)=0$

Example 2

If $f(x)=4x-3$, evaluate

i) find $f(1)$

ii) find $f(2)$

iii) find $f(3)$

Solution:

i) $f(1)=4(1)-3$

$f(1)=4-3$

$f(1)=1$

ii) $f(2)=4(2)-3$

$f(2)=8-3$

$f(2)=5$

iii) $f(3)=4(3)-3$

$f(3)=12-3$

$f(3)=9$

Example 3

If $g(x)=x^2-6$, evaluate

i) $g(5)$

ii) $g(4)$

iii) $g(-4)$

Solution:

i) So, we plug in $5$ for x

$g(5)=(5)^2-6$

$g(5)=25-6$

$g(5)=19$

ii) Let's plug in $4$ for x

$g(4)=(4)^2-6$

$g(4)=16-6$

$g(4)=10$

iii) As usual, we insert $-4$ for x

$g(-4)=(-4)^2-6$

$g(-4)=16-6$

$g(-4)=10$

Example 4

Evaluate $f(x)=2x+3$ when

i) $f(2)$

ii) $f(a)$

iii) $f(a+h)$

iv) $f(a)-f(a+h)$

Solution:

i) $f(2)=2(2)+3$

$f(2)=4+3$

$f(2)=7$

ii) $f(a)$ means we should insert a for x

$f(a)=2(a)+3$

$f(a)=2a+3$

iii) $f(a+h)$ means we should substitute a+h for x

$f(a+h)=2(a+h)+3$

$f(a+h)=2a+2h+3$

iv) $f(a)-f(a+h)$ means we should subtract the output value of $f(a)$ and $f(a+h)$. We know from ii) that $f(a)$ is $2a+3$. We also see from iii) that $f(a+h)$ is $2a+2h+3$

Therefore,

$f(a)-f(a+h)=2a+3-(2a+2h+3)$

Distributing the minus sign

$f(a)-f(a+h)=2a+3-2a-2h-3$

$f(a)-f(a+h)=-2h$

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Example 5

If $g(x)=-x^2+3x+6$, find

i) $g(2+a)$

ii) $g(2+a)-g(2)$

iii) $\frac{g(2+a)-g(2)}{a}$

Solution:

i) $g(2+a)=-(2+a)^2+3(2+a)+6$

Expanding the brackets

$g(2+a)=-(2(2+a)+a(2+a))+3(2+a)+6$

$g(2+a)=-(4+2a+2a+a^2)+6+3a+6$

$g(2+a)=-(4+4a+a^2)+6+3a+6$

$g(2+a)=-4-4a-a^2+6+3a+6$

$g(2+a)=-4+6+6-4a+3a-a^2$

$g(2+a)=8-a-a^2$

ii) As you just from (i), $g(2+a)=8-a-a^2$,

Now, let evaluate $g(2)$

$g(2)=-(2)^2+3(2)+6$

$g(2)=-4+6+6$

$g(2)=8$

Therefore

 $g(2+a)-g(2)=8-a-a^2-8$

$g(2+a)-g(2)=-a-a^2$

iii) since $g(2+a)-g(2)=-a-a^2$

Then,

$\frac{g(2+a)-g(2)}{a}=\frac{-a-a^2}{a}$

$\frac{g(2+a)-g(2)}{a}=\frac{\require{cancel}\bcancel{a}(-1-a)}{\require{cancel}\bcancel{a}}$

$\frac{g(2+a)-g(2)}{a}=-1-a$

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