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