Evaluating Functions

Evaluating Functions

To evaluate a function is to:

Replace (substitute) any variable with its given number or expression

Like in this example:

Example: evaluate the function f(x) = 2x+4 for x=5

Just replace the variable "x" with "5":

f(5) = 2×5 + 4 = 14

Answer: f(5) = 14

More Examples

Here is a function:

f(x) = 1 − x + x2

f is just a name,
x is just a place-holder.

These are all the same function:

  • f(x) = 1 − x + x2
  • f(q) = 1 − q + q2
  • w(A) = 1 − A + A2
  • pumpkin(θ) = 1 − θ + θ2

 

Evaluate For a Given Value:

Let us evaluate that function for x=3:

f(3) = 1 − 3 + 32 = 1 − 3 + 9 = 7

Evaluate For a Given Expression:

Evaluating can also mean replacing with an expression (such as 3m+1 or v2).

Let us evaluate our function for x=1/r:

f(1/r) = 1 − (1/r) + (1/r)2

Or for x=a−4:

f(a−4) = 1 − (a−4) + (a−4)2
  = 1 − a + 4 + a2 − 8a + 16
  = 21 − 9a + a2

Another Example

You can use your ability to evaluate functions in other way:

Example: h(x) = 3x2 + ax − 1

  • You are told that h(3) = 8, can you work out what "a" is?

 

First, evaluate h(3):h(3) = 3×(3)2 + a×3 − 1
Simplify:h(3) = 27 + 3a − 1
 h(3) = 26 + 3a

 

Now ... we are told that h(3) = 8: 8 = 26 + 3a
Swap sides:26 + 3a = 8
Subtract 26 from both sides:3a = −18
Divide by 3:a = −6

 

Check: h(3) = 3(3)2 − 6×3 − 1 = 27 − 18 − 1 = 8 yes

Careful!

I recommend putting the substituted values inside parentheses () , so you don't make mistakes.

Example: evaluate the function h(x) = x2 + 2 for x = −3

Replace the variable "x" with "−3":

h(−3) = (−3)2 + 2 = 9 + 2 = 11

Without the () you could make a mistake:

h(−3) = −32 + 2 = −9 + 2 = −7 (WRONG!)

Also be careful of this:

f(x+a) is not the same as f(x) + f(a)

Example: g(x) = x2

      g(w+1) =(w+1)2
=w2 + 2w + 1

vs

g(w) + g(1) =w2 + 12
=w2 + 1        

Different Result!

 

533, 534, 541, 2431, 542, 1177, 2432, 2433, 1178, 3241