# Evaluate Functions

Evaluate real valued functions: A step by step tutorial, with examples and detailed solutions. To find the value \( f(\text{a}) \) of a function, \(\text{a}\) has to be in the domain of \( f \). In what follows, we are considering only real valued functions.

## Evaluate Functions; Examples with Solutions

### Example 1

Evaluate function \( f \) for \( x = -2 \) and \( x = 2 \), if possible, given that \( f \) is defined by
\[ f(x) = \dfrac{{-4}}{{x + 2}} \]
**Solution to Example 1**

Function \( f \) given above has domain

\( (-\infty, -2) \cup (-2, +\infty) \)

Since at \( x = -2 \) the denominator of \( f(x) \) is equal to 0,

\( f(-2) = \text{undefined} \)

To find \( f(2) \), substitute \( x \) by 2 in \( f(x) = -4 / (x + 2) \)

\( f(2) = \dfrac{{-4}}{{2 + 2}} = -1 \)

### Example 2

Evaluate function \( g \) for \( x = 3 \) and \( x = 0 \), if possible, given that function \( g \) is defined by
\[ g(x) = \sqrt{x - 3} \]
**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

\( [3, +\infty) \)

\( x = 0 \) is not included in the domain, hence

\( g(0) = \sqrt{0 - 3} = \sqrt{-3} = \text{not a real number} \)

### Example 3

Evaluate, if possible, \( h(4) \), \( g(4) \), and \( \dfrac{{h(4)}}{{g(4)}} \) where functions \( h \) and \( g \) are defined by
\[ h(x) = 3x - 8 \quad , \quad g(x) = x^2 - 16 \]
**Solution to Example 3**

Evaluate \( h(4) \)

\( h(4) = 3(4) - 8 = 4 \)

Evaluate \( g(4) \)

\( g(4) = 4^2 - 16 = 0 \)

In evaluating \( \dfrac{{h(4)}}{{g(4)}} \), \( g(4) \) which is the denominator is equal to 0. In mathematics division by zero is not allowed. Hence

\( \dfrac{{h(4)}}{{g(4)}} = \text{undefined} \)

### Example 4

Evaluate, if possible, \( h(t - 1) \) where function \( h \) is defined by
\[ h(x) = 2x^2 - 2x + 2 \]
**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 function \( f \) for \( x = 9 \) given that \( f(x) = 2x^2 + 2 \)

2 - Evaluate \( g(1) \), \( h(1) \), and \( \dfrac{{g(1)}}{{h(1)}} \) given that \( g(x) = x^3 + 1 \) and \( h(x) = x - 1 \)

3 - Evaluate \( f(t + 2) \) given that \( f(x) = -2x^2 + 2x \)
### Solutions to the Above Exercises

1 - \( f(9) = 164 \)

2 - \( g(1) = 2 \), \( h(1) = 0 \), \( \dfrac{{g(1)}}{{h(1)}} = \text{undefined} \)

3 - \( f(t + 2) = -2(t + 2)^2 + 2(t + 2) \)

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