To find the value f(a) of a function, a has to be in the domain of f. In what follows, we are considering only real valued functions.
Example 1: Evaluate, if possible, f(2) and f(2) given that f is defined by
f (x) =  4 / ( x + 2)
Solution to Example 1
 Function f given above has domain
(infinity , 2) U (2 , +infinity)
 Since at x = 2 the denominator of f(x) is equal to 0,
f(2) = undefined.
 To find f(2), substitute x by 2 in f(x) = 4 / ( x + 2)
f(2) =  4 / (2 + 2) = 1.
Example 2: Evaluate, if possible, g(3) and g(0) given that g is defined by
g (x) = sqrt(x  3) , sqrt means square root.
Solution to Example 2
 To find g(3), substitute x by 3 in the formula of the function
g (3) = sqrt(3  3) = sqrt(0) = 0
 The domain of g is given by the interval
[3 , +infinity)
 x = 0 is not included in the domain, hence
g(0) = sqrt(0  3) = sqrt(3) = not a real number.
Example 3: Evaluate, if possible, h(4), g(4) and h(4) / g(4) where functions h and g are defined by
h (x) = 3x  8 , g (x) = x^{ 2}  16
Solution to Example 3
 Evaluate h(4)
h(4) = 3(4)  8 = 4
 Evaluate g(4)
g (4) = 4^{ 2}  16 = 16 16 = 0
 In evaluating h(4) / g(4), g(4) which is the denominator is equal to 0. In mathematics division by zero is not allowed. Hence
h(4) / g(4) = undefined
Example 4: Evaluate, if possible, h(t 1) where function h is defined by
h (x) = 2 x^{ 2}  2 x + 2
Solution to Example 4
 The domain of this function is the set of
all real numbers. Hence h(t 1) is given by
h (t  1) = 2 (t  1)^{ 2}  2 (t  1) + 2
 Expand the square and group like terms
h (t  1) = 2 (t^{ 2}  2t + 1)  2t + 2 + 2
= 2t^{ 2}  4t + 2  2t + 4
= 2t^{ 2}  6t + 6
Exercises:
1  Evaluate f(9) given that f(x) = 2 x^{ 2} + 2
2  Evaluate g(1), h(1) and g(1) / h(1) given that g(x) = x^{ 3} + 1 and h(x) = x  1
3  Evaluate f(t + 2) given that f(x) =  2 x^{ 2} + 2x
Solutions to Above Exercises:
1  f(9) = 164
2  g(1) = 2 , h(1) = 0 , g(1) / h(1) = undefined
3  f(t + 2) =  2 t^{ 2}  6t  4
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