# Evaluating Functions Examples

The evaluating functions examples on this page aim to show how to evaluate functions in Math effectively.

Evaluating functions amounts to substituting a value of some sort into a function in place of a variable and obtaining the result.

We can look at the function    f(x) \space = \space 2x + 4.

If we were asked to evaluate this function at   x \space = \space 3,
we would just plug this value into the function.

f(3) \space = \space 2(3) + 4 \space = \space 6 + 4 \space = \space 10

This process follows on in other evaluating functions examples also.

### Evaluating Functions Examples:

(1.1)

Evaluate    f(x) = \space 3x \space {\text{–}} \space 11     at     x = 3.

Solution

f(3) = \space 3(3) \space {\text{–}} \space 11 \space = \space 9 \space {\text{–}} \space 11 \space = \space {\text{-}}2

(1.2)

Evaluate    f(b) = \space 4 + b + b^{2}     at     b = 2.

Solution

f(2) = \space 4 + 2 + (2)^{2} \space = \space 4 + 2 + 4 \space = \space 10

(1.3)

Evaluate    h(x) = \space {\frac{8}{x}} + 1     at     x = 4.

Solution

h(4) = \space {\frac{8}{4}} + 1 \space = \space 2 + 1 \space = \space 3

## Further Evaluating Functions Examples

It’s not only numbers that can be used in functions when evaluating them.

Expressions can also be input into functions to obtain a result.

(2.1)

Evaluate    f(x) = \space 6x + 4     for     x = {\frac{t}{3}}.

Solution

f({\frac{t}{3}}) = \space 6({\frac{t}{3}}) + 4 \space = \space {\frac{6t}{3}} + 4 \space = \space 2t + 4

(2.2)

Evaluate    g(b) = \space 5 + b^{2}     for     b = x \space {\text{–}} \space 2.

Solution

g(x \space {\text{–}} \space 2) \space = \space 5 + (x \space {\text{–}} \space 2)^{2}

= \space 5 + x^2 \space {\text{–}} \space 4x + 4 \space = \space x^2 \space {\text{–}} \space 4x + 9

(2.3)

f(v) = \space bv \space {\text{–}} \space 2v^{2} + 1.

If   f(2) = 3,   what is the value of  b?

Solution

The first step is to evaluate f(2).

f(2) = \space 2b \space {\text{–}} \space 2(2)^{2} + 1 \space = \space 2b \space {\text{–}} \space 8 + 1 \space = \space 2b \space {\text{–}} \space 7

So we know   2b \space {\text{–}} \space 7 = 3,  we can proceed from here to obtain  b.

2b \space {\text{–}} \space 7 + 7 \space = \space 3 + 7     =>     2b = 10

\frac{2b}{2} \space = \space \frac{10}{2}     =>     b = 5

We can check by evaluating with our found  b  value.

f(2) = \space 5(2) \space {\text{–}} \space 2(2)^{2} + 1 \space = \space 10 \space {\text{–}} \space 8 + 1 \space = \space 3

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