Question

1. state the domain and range of each of the following functions (1 point each):

a. $f(x)=\frac{1}{x + 3}$
b. $g(x)=sqrt{x+6}$
c. $h(x)=x^{2}+2x + 5$

Explanation:

Step1: Find domain of \(f(x)=\frac{1}{x + 3}\)

The denominator cannot be zero. So \(x+3
eq0\), which gives \(x
eq - 3\). Domain is \(\{x\in\mathbb{R}:x
eq - 3\}\).

Step2: Find range of \(f(x)=\frac{1}{x + 3}\)

Let \(y = \frac{1}{x+3}\), then \(x+3=\frac{1}{y}\), \(y
eq0\). Range is \(\{y\in\mathbb{R}:y
eq0\}\).

Step3: Find domain of \(g(x)=\sqrt{x + 6}\)

The expression under the square - root must be non - negative. So \(x+6\geq0\), which gives \(x\geq - 6\). Domain is \(\{x\in\mathbb{R}:x\geq - 6\}\).

Step4: Find range of \(g(x)=\sqrt{x + 6}\)

Since the square root of a non - negative number is non - negative, \(y=\sqrt{x + 6}\geq0\). Range is \(\{y\in\mathbb{R}:y\geq0\}\).

Step5: Find domain of \(h(x)=x^{2}+2x + 5\)

It is a polynomial function. The domain of any polynomial function is all real numbers. Domain is \(\{x\in\mathbb{R}\}\).

Step6: Find range of \(h(x)=x^{2}+2x + 5\)

Complete the square: \(h(x)=(x + 1)^{2}+4\). Since \((x + 1)^{2}\geq0\), then \((x + 1)^{2}+4\geq4\). Range is \(\{y\in\mathbb{R}:y\geq4\}\).

Answer:

a. Domain: \(\{x\in\mathbb{R}:x
eq - 3\}\), Range: \(\{y\in\mathbb{R}:y
eq0\}\)
b. Domain: \(\{x\in\mathbb{R}:x\geq - 6\}\), Range: \(\{y\in\mathbb{R}:y\geq0\}\)
c. Domain: \(\{x\in\mathbb{R}\}\), Range: \(\{y\in\mathbb{R}:y\geq4\}\)