Function range

Find the range of a function

Example. Find the range of \(y = 2 + \frac{x^2}{x^2 + 4}\)

Solution. First find the domain of the function: all real numbers, or \((-\infty, +\infty)\). Then step-by-step work on the independent variable:

$$ -\infty \lt x \lt \infty $$

$$ \Rightarrow \quad 0 \le x^2 \lt \infty $$

$$ \Rightarrow \quad 4 \le x^2 + 4 \lt \infty $$

$$ \Rightarrow \quad 0 \lt \frac{1}{x^2 + 4} \le \frac 14 $$

$$ \Rightarrow \quad 0 \le \frac{x^2}{x^2 + 4} \lt 1 $$

$$ \Rightarrow \quad 2 \le 2 + \frac{x^2}{x^2 + 4} \lt 3 $$

Therefore, the range is the interval \([2,3)\)

Find the range of a composite function

Example. Find a formula for \(f \circ g\) and find domain and range of it.

$$ f(x) = \sqrt{x + 1}, \quad g(x) = \frac 1x $$

Solution.

$$ f \circ g = \sqrt{\frac 1x + 1} = \sqrt{\frac{x + 1}{x}} $$

In order to find the domain of \(f \circ g\), we need to find an interval where \(\frac{x + 1}{x} \ge 0\) which results in the interval \((-\infty, -1] \cup (0, \infty)\)

To find the range of \(f \circ g\), we need to find the value of function on two separate intervals from the domain.

$$ -\infty \lt x \le -1 $$

$$ \Rightarrow -1 \le \frac 1x \lt 0 $$

$$ \Rightarrow 0 \le \frac 1x + 1 \lt 1 $$

$$ \Rightarrow 0 \le \sqrt{\frac 1x + 1} \lt 1 $$

$$ 0 \lt x \lt \infty $$

$$ 0 \lt \frac 1x \lt \infty $$

$$ 1 \lt \frac 1x + 1 \lt \infty $$

$$ 1 \lt \sqrt{\frac 1x + 1} \lt \infty $$

The range of \(f \circ g\) is the interval \([0,1) \cup (1,\infty)\)