# 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 - $$f(t + 2) = -2(t + 2)^2 + 2(t + 2)$$
More mathematics tutorials and problems are presented on this site.