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by: G. Martin ;
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When evaluating a function, you
are simply finding the value of f(x)
given the value of x. It means that
we plug the value of x into the
function.
Example 1. Evaluate f(x) = 2x – 3
when x = 2
f(x) = 2x – 3
f(2) = 2(2) – 3
= 4 – 3
f(2) = 1
Thus f(x) = 2x – 3 when x – 2 is 1.
Example 2. Evaluate f(x) = 3x + 7
when x = -1
f(x) = 3x + 7
f(-1) = 3(-1) + 7
= -3 + 7
f(-1) = 4
Thus f(x) = 3x + 7 when x = -1 is 4.
Example 3. Evaluate f(x) = x2 + x – 2
when x = 5
f(x) = x2 + x – 2
f(5) = (5)2 + 5 - 2
= 25 + 5 – 2
= 30 – 2
f(5) = 28
Thus (x) x2 + x -2 when x = 5 is 28.
Example 4. Evaluate g(x) = x2 – x - 5
when x = -3
g(x) = x2 – x - 5
g(-3) = (-3)2 – (-3) – 5
= 9 + 3 – 5
= 12 – 5
g(x) = 7
Thus g(x) = x2 – x – 5 when x = -3 is 7.
Example 5. Complete the function table for f(x) = 3x2 – 5x + 10, given the
values for x are {-1, 0, 1, 2, 3}
Input f(x) Rule: 3x2 – 5x + 10 Output f(x)
-1 3(-2)2 – 5(-1) + 10 = 3 + 5 + 10 = 18 18
0 3(0)2 – 5(0) + 10 = 0 – 0 + 10 = 10 10
1 3(1)2 – 5 (1) + 10 = 3 – 5 + 10 = 8 8
2 3(2)2 – 5(2) + 10 = 12 – 10 + 10 = 12 12
3 3(3)2 – 5(3) + 10 = 27 – 15 + 10 = 22 22

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Evaluating a function

  • 2. Lets Learn! When evaluating a function, you are simply finding the value of f(x) given the value of x. It means that we plug the value of x into the function.
  • 3. Example 1. Evaluate f(x) = 2x – 3 when x = 2 f(x) = 2x – 3 f(2) = 2(2) – 3 = 4 – 3 f(2) = 1 Thus f(x) = 2x – 3 when x – 2 is 1. Example 2. Evaluate f(x) = 3x + 7 when x = -1 f(x) = 3x + 7 f(-1) = 3(-1) + 7 = -3 + 7 f(-1) = 4 Thus f(x) = 3x + 7 when x = -1 is 4.
  • 4. Example 3. Evaluate f(x) = x2 + x – 2 when x = 5 f(x) = x2 + x – 2 f(5) = (5)2 + 5 - 2 = 25 + 5 – 2 = 30 – 2 f(5) = 28 Thus (x) x2 + x -2 when x = 5 is 28. Example 4. Evaluate g(x) = x2 – x - 5 when x = -3 g(x) = x2 – x - 5 g(-3) = (-3)2 – (-3) – 5 = 9 + 3 – 5 = 12 – 5 g(x) = 7 Thus g(x) = x2 – x – 5 when x = -3 is 7.
  • 5. Example 5. Complete the function table for f(x) = 3x2 – 5x + 10, given the values for x are {-1, 0, 1, 2, 3} Input f(x) Rule: 3x2 – 5x + 10 Output f(x) -1 3(-2)2 – 5(-1) + 10 = 3 + 5 + 10 = 18 18 0 3(0)2 – 5(0) + 10 = 0 – 0 + 10 = 10 10 1 3(1)2 – 5 (1) + 10 = 3 – 5 + 10 = 8 8 2 3(2)2 – 5(2) + 10 = 12 – 10 + 10 = 12 12 3 3(3)2 – 5(3) + 10 = 27 – 15 + 10 = 22 22