# Find Range of Functions

Find the range of real valued mathematical functions using different techniques.

## Examples with Solutions

### Example 1

Find the Range of function $$f$$ defined by $f(x) = -3$

#### Solution to Example 1

The given function has a constant value 3 and therefore the range is the set
$$\{3\}$$

### Example 2

Find the Range of function $$f$$ defined by $f(x) = 4x + 5$

#### Solution to Example 2

Assuming that the domain of the given function is the set of all real numbers $$\mathbb{R}$$, so that the variable $$x$$ takes all values in the interval
$$(-\infty , +\infty)$$
If $$x$$ takes all values in the interval $$(-\infty , +\infty)$$ then $$4x + 5$$ takes all values in the interval $$(-\infty , +\infty)$$ and the range of the given function is given by the interval
$$(-\infty , +\infty)$$

### Example 3

Find the Range of function $$f$$ defined by $f(x) = x^2 + 5$

#### Solution to Example 3

Assuming that the domain of the given function is $$\mathbb{R}$$ meaning that $$x$$ takes all values in the interval $$(-\infty , +\infty)$$ which means that $$x^2$$ is either zero or positive. Hence we can write the following inequality
$$x^2 \geq 0$$
Add 5 to both sides of the inequality to obtain the inequality
$$x^2 + 5 \geq 5$$ or $$f(x) \geq 5$$
The range of $$f(x) = x^2 + 5$$ is given by the interval: $$[5 , +\infty)$$

### Example 4

Find the Range of function $$f$$ defined by $f(x) = -2x^2 + 4x - 7$

#### Solution to Example 4

We first write the given quadratic function in vertex form by completing the square
$$f(x) = -2x^2 + 4x - 7 = -2(x^2 - 2x) - 7 = -2( (x - 1)^2 - 1) - 7 = -2(x - 1)^2 - 5$$
The domain of the given function is $$\mathbb{R}$$ with $$x$$ taking any value in the interval $$(-\infty , +\infty)$$ hence $$(x - 1)^2$$ is either zero or positive. We start by writing the inequality
$$(x - 1)^2 \geq 0$$
Multiply both sides of the inequality by -2 and change the symbol of inequality to obtain
$$-2 (x - 1)^2 \leq 0$$
Add $$-5$$ to both sides of the inequality to obtain
$$-2 (x - 1)^2 - 5 \leq -5$$ or $$f(x) \leq -5$$ and hence the range of function $$f$$ is given by the interval $$(-\infty , -5]$$

### Example 5

Find the Range of function $$f$$ defined by $f(x) = -3e^{2x + 5} + 2$

#### Solution to Example 5

The domain of the given function is $$\mathbb{R}$$ and therefore $$x$$ takes all values in the interval $$(-\infty , +\infty)$$. The exponent $$2x + 5$$ takes all values in the interval $$(-\infty , +\infty)$$, we can therefore write the following inequality (basic exponential function always positive)
$$e^{2x + 5} > 0$$
Multiply both sides of the inequality by -3 and change symbol of inequality to obtain
$$-3e^{2x + 5} < 0$$
Add 2 to both sides of the inequality to obtain
$$-3e^{2x + 5} + 2 \lt 2$$ or $$f(x) \lt 2$$
The range of the given function is given by the interval $$(-\infty , 2)$$.

### Example 6

Find the Range of function $$f$$ defined by $f(x) = -3e^{x^2 + 5} + 2$

#### Solution to Example 6

The domain of the given function is $$\mathbb{R}$$, we can start by writing the inequality
$$x^2 \geq 0$$
Add 5 to both sides of the inequality
$$x^2 + 5 \geq 5$$
The basic exponential function being an increasing function, we can use the above to write the inequality
$$e^{x^2 + 5} \geq e^5$$
Multiply both sides of the inequality by -3 and add 2 to both sides to obtain
$$-3e^{x^2 + 5} + 2 \leq -3e^5 + 2$$
Note that the left hand side of the inequality is equal to $$f(x)$$. Hence
$$f(x) \lt -3e^5 + 2$$ which means that the range of function $$f$$ is given by the interval: $$(-\infty , -3e^5 + 2)$$

### Example 7

Find the Range of function $$f$$ defined by $f(x) = \dfrac{x - 1}{x + 2}$

#### Solution to Example 7

For this rational function, a direct algebraic method similar to those above is not obvious. Let us first find its inverse, the domain of its inverse which give the range of $$f$$.
We first prove that $$f$$ is a one to one function and then find its inverse. For a function to be a one to one, we need to show that
If $$f(a) = f(b)$$ then $$a = b$$.
$$\dfrac{a - 1}{a + 2} = \dfrac{b - 1}{b + 2}$$
Cross multiply, expand and simplify
$$(a - 1)(b + 2) = (b - 1)(a + 2)$$
$$ab + 2a - b - 2 = ab + 2b - a - 2$$
$$3a = 3b$$, which finally gives $$a = b$$ and proves that $$f$$ is a one to one function.
Let us find the inverse of $$f$$
$$y = \dfrac{x - 1}{x + 2}$$
Solve for $$x$$
$$x = \dfrac{2y + 1}{1 - y}$$
change $$y$$ into $$x$$ and $$x$$ into $$y$$ and write the inverse function
$$f^{-1}(x) = y = \dfrac{2x + 1}{1 - x}$$
The range of $$f$$ is given by the domain of $$f^{-1}$$ and is therefore given by the interval
$$(-\infty , 1) \cup (1 , +\infty)$$

### Example 8

Find the Range of function $$f$$ defined by $f(x) = -3\ln(x + 3) - 2$

#### Solution to Example 8

The range of $$\ln(x)$$ is given by the interval $$(-\infty , +\infty)$$. Since the graph of $$\ln(x + 3)$$ is the graph of $$\ln(x)$$ shifted 3 units to the left, the range of $$\ln(x + 3)$$ is also given by the interval $$(-\infty , +\infty)$$.
The graph of $$-3\ln(x + 3)$$ is that of $$\ln(x + 3)$$ reflected on the x-axis because of the minus sign and expanded vertically by 3 and therefore the range is still given by the interval $$(-\infty , +\infty)$$.
The graph of $$f$$ is that of $$-3\ln(x +3)$$ shifted 2 units up and therefore the range is also given by the interval $$(-\infty , +\infty)$$.

### Example 9

Find the Range of function $$f$$ defined by $f(x) = \dfrac{1}{x^2 + 4}$

#### Solution to Example 9

$$x^2$$ is a quantity which may be zero or positive and therefore we can write
$$x^2 \geq 0$$
add 4 to both sides of the inequality to obtain
$$x^2 + 4 \geq 4$$
Divide both side of the inequality $$x^2 + 4 \geq 4$$ by the positive quantity $$4(x^2 + 4)$$ to obtain
$$\dfrac{1}{4} \geq \dfrac{1}{x^2 + 4}$$
The right side of the inequality is equal to $$f(x)$$. Hence the above inequality may be written as
$$f(x) \leq \dfrac{1}{4}$$
Because $$\dfrac{1}{x^2 + 4}$$ is always positive and never zero but may be very close to zero as $$x$$ increases or decreases indefinitely , the range of $$f$$ is given by the interval $$(0; \dfrac{1}{4})$$

### Example 10

Find the Range of function $$f$$ defined by $f(x) = \sqrt{x^2 + 4x + 4} - 6$

#### Solution to Example 10

Note that $$x^2 + 4x + 4 = (x + 2)^2$$. Therefore
$$f(x) = \sqrt{x^2 + 4x + 4} - 6 = \sqrt{(x + 2)^2} - 6 = |x + 2| - 6$$
The range of $$|x + 2|$$ is given by the interval $$[0 , +\infty)$$. The graph of $$f$$ is that of $$|x + 2|$$ shifted down by 6 units and therefore the range of is given by the interval $$[-6 , +\infty)$$

### Example 11

Find the Range of function $$f$$ defined by $f(x) = -2\sin(3x - \pi) + 1.5$

#### Solution to Example 11

From basic trigonometry we know that the range of values of sine function is $$[-1 , 1]$$. Hence
$$-1 \leq \sin(3x - \pi) \leq 1$$
Multiply all terms by -2; change symbols of inequalities and add 1.5 to all terms
$$2 + 1.5 \geq -2\sin(3x - \pi) + 1.5 \geq -2 + 1.5$$
Simplify and rewrite as
$$-0.5 \leq f(x) \leq 3.5$$ which gives the range of $$f$$ as the interval $$[-0.5 , 3.5]$$.

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