Question

⠀
Q28. Find the domain and range of $\sf{\sqrt{x^2-16}}.$​

Answer 1

Answer:

Answer

Consider the given function.

f(x)=

16−x

2

For domain under root should not be negative quantity,

16−x

2

≥0

16≥x

2

Therefore,

x≤4 or x≥−4

Then,

The domain[−4,4]

Range: f(x) is maximum at x=0,f(x)=4

And f(x) is minimum at x=4,f(x)=0

Range

[0,4]

Hence, this is the answer.

hope it helps you

Answer 2

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:f(x) = \sqrt{ {x}^{2} - 16}$

Domain

We know, Domain of a function is defined as the set of those values where function is well defined.

So,

$\rm :\longmapsto\:f(x) = \sqrt{ {x}^{2} - 16} \: is \: defined \: when \:$

$\rm :\longmapsto\: {x}^{2} - 16 \geqslant 0$

$\rm :\longmapsto\: {x}^{2} - {4}^{2} \geqslant 0$

$\rm :\longmapsto\:(x - 4)(x + 4) \geqslant 0$

$\rm :\implies\:x \leqslant - 4 \: \: or \: \: x \geqslant 4$

$\bf\implies \:x \: \in \: ( - \infty , - 4] \: \cup \: [4, \infty )$

Range

We know, Range of a function is defined as set of those values taken by f(x) at the points in its domain.

Now,

Given function is

$\rm :\longmapsto\:f(x) = \sqrt{ {x}^{2} - 16 }$

Consider,

Case :- 1

$\red{\rm :\longmapsto\:When \: x \geqslant 4}$

$\rm :\longmapsto\: {x}^{2} - 16 \geqslant 0$

$\rm :\implies\: \sqrt{ {x}^{2} - 16} \geqslant 0$

$\bf\implies \:f(x) \geqslant 0$

$\bf\implies \:f(x) \: \in \: [0, \: \infty )$

Now,

Case :- 2

$\red{\rm :\longmapsto\:When \: x \leqslant 4}$

$\rm :\longmapsto\: {x}^{2} - 16 \geqslant 0$

$\rm :\implies\: \sqrt{ {x}^{2} - 16} \geqslant 0$

$\bf\implies \:f(x) \geqslant 0$

$\bf\implies \:f(x) \: \in \: [0, \: \infty )$

Hence,

Range of f(x) is

$\bf\implies \:f(x) \: \in \: [0, \: \infty )$

$\begin{gathered}\begin{gathered}\rm :\longmapsto\:\bf\: Hence-\begin{cases} &\sf{Domain: \: x \in \: ( - \infty , - 4] \cup \: [4, \infty )} \\ \\ &\sf{Range : \: f(x) \in \: [0, \infty ) } \end{cases}\end{gathered}\end{gathered}$

Additional Information :-

If a > b, then

$\boxed{ \bf{ \: (x - b)(x - a) < 0\bf\implies \:b < x < a}}$

$\boxed{ \bf{ \: (x - b)(x - a) \leqslant 0\bf\implies \:b \leqslant x \leqslant a}}$

$\boxed{ \bf{ \: (x - b)(x - a) > 0\bf\implies \:x < b \: \: or \: \: x > a}}$

$\boxed{ \bf{ \: (x - b)(x - a) \geqslant 0\bf\implies \:x \leqslant b \: \: or \: \: x \geqslant a}}$

$\boxed{ \bf{ \: |x| < y \: \bf\implies \: - y < x < y}}$

$\boxed{ \bf{ \: |x| \leqslant y \: \bf\implies \: - y \leqslant x \leqslant y}}$

$\boxed{ \bf{ \: |x| > y \: \bf\implies \:x < - y \: \: or \: \: x > y}}$

$\boxed{ \bf{ \: |x| \geqslant y \: \bf\implies \:x \leqslant - y \: \: or \: \: x \geqslant y}}$