# Evaluate Mathematical Functions

Evaluate real valued functions: A step by step tutorial, with examples and detailed solutions. 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.

## Examples with Solutions## Example 1Evaluate, if possible, f(-2) and f(2) given that f is defined bySolution to Example 1Function 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 2Evaluate, if possible, g(3) and g(0) given that g is defined bySolution to Example 2To find g(3), substitute x by 3 in the formula of the function g (3) = √(3 - 3) = √(0) = 0 The domain of g is given by the interval [3 , +infinity) x = 0 is not included in the domain, hence g(0) = √(0 - 3) = √(-3) = not a real number.
## Example 3Evaluate, if possible, h(4), g(4) and h(4) / g(4) where functions h and g are defined by^{ 2} - 16
Solution to Example 3Evaluate 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 4Evaluate, if possible, h(t -1) where function h is defined by^{ 2} - 2 x + 2
Solution to Example 4The 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
## Exercises1 - 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 the Above Exercises1 - 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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