Question

Find Domain and and range:

f(x)=√x²-16​

Answer 1

Answer:

Domain of f(x): x≤ -4 or x≥4

Range is [0,∞) or f(x)≥0


Solution:

To find the domain of the function,put the expression under the square root with the condition greater than and equal to zero.

Since we know that square root does not give real values when it was on negative values.

So,

$\sqrt{ {x}^{2} - 16 } \geqslant 0 \\ \\ {x}^{2} - 16 \geqslant 0 \\ \\ {x}^{2} \geqslant 16 \\ \\ x \geqslant \sqrt{16} \\ \\ so \\ \\ x \leqslant - 4 \\ \\ or \\ \\ x \geqslant 4$
Domain of f(x): x≤ -4 or x≥4

Range:

To find the range,put y=f(x)

$y = \sqrt{ {x}^{2} - 16 } \\ \\ {y}^{2} = {x }^{2} - 16 \\ \\ {y}^{2} + 16 = {x}^{2} \\ \\ x = \sqrt{ {y}^{2} + 16}$
since for all values of y,their exist a x.

So range is [0,∞)

or f(x)≥0

Answer 2

Answer:

Domain of f(x): x≤ -4 or x≥4

Range is [0,∞) or f(x)≥0

Solution:

To find the domain of the function,put the expression under the square root with the condition greater than and equal to zero.

Since we know that square root does not give real values when it was on negative values.

So,

\begin{gathered} \sqrt{ {x}^{2} - 16 } \geqslant 0 \\ \\ {x}^{2} - 16 \geqslant 0 \\ \\ {x}^{2} \geqslant 16 \\ \\ x \geqslant \sqrt{16} \\ \\ so \\ \\ x \leqslant - 4 \\ \\ or \\ \\ x \geqslant 4 \\ \\\end{gathered}

x

2

−16

⩾0

x

2

−16⩾0

x

2

⩾16

x⩾

16

so

x⩽−4

or

x⩾4

Domain of f(x): x≤ -4 or x≥4

Range:

To find the range,put y=f(x)

\begin{gathered}y = \sqrt{ {x}^{2} - 16 } \\ \\ {y}^{2} = {x }^{2} - 16 \\ \\ {y}^{2} + 16 = {x}^{2} \\ \\ x = \sqrt{ {y}^{2} + 16} \\ \\ \end{gathered}

y=

x

2

−16

y

2

=x

2

−16

y

2

+16=x

2

x=

y

2

+16

since for all values of y,their exist a x.

So range is [0,∞)

or f(x)≥0