# Lesson 2

Lesson 2: Evaluating Functions
To evaluate a function,
1.Substitute the value in function of x.
2.Replace the all the variables with the number or expression for the func- tion’s variable.
3.Evaluate and simplify the function.
Example 1: Given the function: f(x ) = 2 x+ 1, nd f(6).
Substitute 6 in place holder x,
f(6) = 2 x+ 1
Replace the all the variables with 6,
f(6) = 2(6) + 1
Then evaluate and simplify the function. f(6) = 12 + 1
f (6) = 13
Therefore, f(6) = 13. It can also write in ordered pair which is (6,13).
Example 2:
Given the function f(x ) = x2
+ 2 x+ 4 when x= 4.
Substitute -4 in the place holder x,
f ( 4) = x2
+ 2 x+ 4
Replace the all the variables with 6, f( 4) = ( 4) 2
+ 2( 4) + 4
Then evaluate and simplify the function. f( 4) = (16) + ( 8) + 4
f ( 4) = 12
Therefore, f( 4) = 12 or simply as ( 4;12) :
Example 3:
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Given
g(x ) = x2
+ 2 x- 1. Find g(2y).
g(2 y) = x2
+ 2 x 1
g (2 y) = (2 y)2
+ 2(2 y) 1
g (2 y) = 4 y2
+ 4 y 1
Therefore, 4( y)2
+ 4 y 1:
Example 4:
Given f(x ) = 2 x2
+ 4 x- 12, nd f(2 x+ 4).
Solution: f(2 x+ 4) = 2 x2
+ 4 x 12
= 2(2 x+ 4) 2
+ 4(2 x+ 4) 12
= 2(2 x+ 4)(2 x+ 4) + 4(2 x+ 4) 12
= 2(4 x2
+ 16 x+ 16) + 4(2 x+ 4) 12
= (8 x2
+ 32 x+ 32) + (8 x+ 16) 12
Combine like terms f(2 x+ 4) = 8 x2
+ (32 x+ 8 x) + (32 + 16 12)
= 8 x2
+ 40 x+ 36
= 2(2 x2
+ 10 x+ 9)
Therefore, f(2 x+ 4) = 2(2 x2
+ 10 x+ 9).
Example 5:
Given f(x ) = x2
-x – 4. If f(a ) = 8, what is the value of a?
Solution: Equate the function by x2
-x 4 = 8
x 2
x 4 = 8
x 2
x 12 = 0
( x 4)( x+ 3) = 0
x 4 + 0; x+ 3 = 0
x = 4; x= 3
Therefore, the value of a can be either 4 or -3.
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Exercises:
Evaluate each function
Given the functions:
1. k(x ) = x + 2, nd k(-5)
2. p(x ) = 2 x, nd p(-4)
3. g(n ) = n2
+ 3, nd g(8)
4. g(x ) = x3
+ 5 x, nd g(5)
5. h(n ) = n3
+ 3 n2
, nd h(-5)
6. w(a ) = a2
+ 5 a, nd w(7)
7. p(a ) = a3
– 4 a, nd p(-6)
8. h(n ) = 4 3
n
+ 8 5
, nd
h(-1)
9. f(x) = -1 + 1 4
x;
nd f3 4
10. h(n) = n3
+ 6 n, nd h(4)
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