Question

Find the domain and range and sketch the graph of the function\(h\left( x \right) = \sqrt {4 - {x^2}} \).

Short answer

The domain of the function \(h\left( x \right) = \sqrt {4 - {x^2}} \)is\(\left( { - 2,\;2} \right)\). The range of the function \(h\left( x \right) = \sqrt {4 - {x^2}} \)is\(\left( {0,\;2} \right)\). The graph of the function \(h\left( x \right) = \sqrt {4 - {x^2}} \) is as follows.

Step-by-step solution

Describe the domain

The domain is the set of all values at which the given function is defined.

Find the domain and range and sketch the graph of the function

The function \(h\left( x \right)\) is defined only when\(4 - {x^2} \ge 0\), since the square root of a negative number is not defined.

The domain of \(h\) consists of all values of \(x\) such that\(4 - {x^2} \ge 0\).

\(\begin{array}{c}4 - {x^2} \ge 0\\{x^2} \le 4\\\left| x \right| \le 2\end{array}\)

This means that\( - 2 \le x \le 2\).

Therefore, the domain of the function \(h\left( x \right) = \sqrt {4 - {x^2}} \) is \( - 2 \le x \le 2\)or\(\left( { - 2,\;2} \right)\).

Substitute the values of the domain interval in\(h\left( x \right) = \sqrt {4 - {x^2}} \), and find the minimum and maximum values.

\(\begin{aligned}h\left( { - 2} \right) &= \sqrt {4 - {{\left( { - 2} \right)}^2}} \\ &= \sqrt {4 - 4} \\ &= 0\end{aligned}\)

And,

\(\begin{aligned}h\left( 0 \right) &= \sqrt {4 - {{\left( 0 \right)}^2}} \\ &= \sqrt 4 \\ &= 2\end{aligned}\)

And,

\(\begin{aligned}h\left( 2 \right) &= \sqrt {4 - {{\left( 2 \right)}^2}} \\ &= \sqrt {4 - 4} \\ &= 0\end{aligned}\)

The minimum value of \(h\left( x \right)\) is \(0\) and the maximum value of \(h\left( x \right)\)is\(2\).

Therefore, the range of the function \(h\left( x \right) = \sqrt {4 - {x^2}} \)is\(\left( {0,\;2} \right)\).

Draw the graph of the function \(h\left( x \right) = \sqrt {4 - {x^2}} \)as follows.