Question

Find the range of the real function defined as:-
${f(x) = \sqrt{2 - x} - \dfrac{ \sqrt{25 - {x}^{2} } }{ \sqrt{ {x}^{2} - 4 } } + |x|}$
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Answer 1

First let us find the domain of the function,

$\small\text{ \longrightarrow f(x)=\sqrt{2-x}-\dfrac{\sqrt{25-x^2}}{\sqrt{x^2-4}}+|x| }$

The restrictions in the equations are,

• $\small2-x\in[0,\ \infty)$ from the term $\small\sqrt{2-x}$   ...(1)

• $\small25-x^2\in[0,\ \infty)$ from the term $\small\sqrt{25-x^2}$   ...(2)

• $\smallx^2-4\in(0,\ \infty)$ from the term $\small\sqrt{x^2-4}$ which is also the denominator of a fraction.   ...(3)

Solving (1) we get,

$\small\longrightarrow x\in(-\infty,\ 2]\quad\dots(i)$

Solving (2) we get,

$\small\longrightarrow x\in[-5,\ 5]\quad\dots(ii)$

Solving (3) we get,

$\small\longrightarrow x\in(-\infty,\ -2)\cup(2,\ \infty)\quad\dots(iii)$

Now taking (i) ∧ (ii) ∧ (iii),

$\small\longrightarrow x\in[-5,\ -2)$

This is the domain of the function. It is clear that x < 0 so |x| = -x in the definition of f(x), i.e.,

$\small\text{ \longrightarrow f(x)=\sqrt{2-x}-\dfrac{\sqrt{25-x^2}}{\sqrt{x^2-4}}-x }$

Taking the first derivative of f(x), we get,

$\small\text{ \longrightarrow f'(x)=-\dfrac{1}{2\sqrt{2-x}}+\dfrac{21x}{(x^2-4)\sqrt{(25-x^2)(x^2-4)}}-1 }$

$\small\text{ \longrightarrow f'(x)=\dfrac{21x}{(x^2-4)\sqrt{(25-x^2)(x^2-4)}}-\left(\dfrac{1}{2\sqrt{2-x}}+1\right) }$

The term $\small\text{ \dfrac{1}{2\sqrt{2-x}}+1 }$ is always positive for every x in the domain. Let $\small\text{ \dfrac{1}{2\sqrt{2-x}}+1=m,\quad m>0. }$

Consider the term $\small\text{ \dfrac{21x}{(x^2-4)\sqrt{(25-x^2)(x^2-4)}}. }$ Here 21 > 0, x < 0 as it's clear from the domain, $\small\sqrt{(25-x^2)(x^2-4)}>0$ (√k is always positive if it's denominator of a fraction) and $\smallx^2-4>0$ from (3).

Only x is negative here and the others are positive. Thus the whole term is negative. Let $\small\text{ \dfrac{21x}{(x^2-4)\sqrt{(25-x^2)(x^2-4)}}=-n,\quad n>0. }$

Now,

$\small\longrightarrow f'(x)=-n-m=-(m+n)$

This means f'(x) is always negative, which implies f(x) is strictly decreasing in the domain.

Hence the range of the function is in the form,

$\small\longrightarrow f(x)\in\big(f(-2),\ f(-5)\big]$

Now,

$\small\text{ \longrightarrow f(-2)=\sqrt{2-(-2)}-\dfrac{\sqrt{25-(-2)^2}}{\sqrt{(-2)^2-4}}-(-2) }$

$\small\text{ \longrightarrow f(-2)=4-\dfrac{\sqrt{21}}{0} }$

$\small\longrightarrow f(-2)\to-\infty$

Since f(x) is a strictly decreasing function, the value of f reduces and reduces so f(-2) must approach -∞. Or if one takes $\smallf(-2)\to\infty$ then it cannot be at left of range interval.

And,

$\small\text{ \longrightarrow f(-5)=\sqrt{2-(-5)}-\dfrac{\sqrt{25-(-5)^2}}{\sqrt{(-5)^2-4}}-(-5) }$

$\small\longrightarrow f(-5)=5+\sqrt{7}$

Hence,

$\small\text{ \longrightarrow\underline{\underline{f(x)\in\left(-\infty,\ 5+\sqrt7\right]}} }$

This is the range of the function.