Question

Find the range of the function $f(x) = x\sqrt{1-x^2}$

Hint: Solve it either by using AM-GM relation, or by using the concept of maxima and minima.​

Answer 1

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

The given function is

$\rm :\longmapsto\:f(x) = x \sqrt{1 - {x}^{2} }$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx} f(x) = \dfrac{d}{dx}x \sqrt{1 - {x}^{2} }$

We know

$\boxed{ \tt{ \: \dfrac{d}{dx}uv \: = \: u\dfrac{d}{dx}v \: + \: v\dfrac{d}{dx}u \: }}$

So, using this, we get

$\rm :\longmapsto\:f'(x) = x\dfrac{d}{dx} \sqrt{1 - {x}^{2} } + \sqrt{1 - {x}^{2} }\dfrac{d}{dx}x$

$\rm :\longmapsto\:f'(x) = x \times \dfrac{1}{2 \sqrt{1 - {x}^{2} } }\dfrac{d}{dx}(1 - {x}^{2}) + \sqrt{1 - {x}^{2} } \times 1$

$\rm :\longmapsto\:f'(x) = \dfrac{x}{2 \sqrt{1 - {x}^{2}}} ( - 2x) + \sqrt{1 - {x}^{2} }$

$\rm :\longmapsto\:f'(x) = - \dfrac{ {x}^{2} }{\sqrt{1 - {x}^{2}}} + \sqrt{1 - {x}^{2} }$

$\rm :\longmapsto\:f'(x) = \dfrac{ - {x}^{2} + 1 - {x}^{2} }{\sqrt{1 - {x}^{2}}}$

$\rm :\longmapsto\:f'(x) = \dfrac{1 - 2{x}^{2} }{\sqrt{1 - {x}^{2}}}$

$\rm :\longmapsto\:f'(x) = \dfrac{(1 - \sqrt{2} x) \: (1 + \sqrt{2} x)}{\sqrt{1 - {x}^{2}}}$

For maxima or minima,

$\rm :\longmapsto\:f'(x) = 0$

$\rm :\longmapsto\:\dfrac{(1 - \sqrt{2} x) \: (1 + \sqrt{2} x)}{\sqrt{1 - {x}^{2}}} = 0$

$\bf\implies \:x = \dfrac{1}{ \sqrt{2} } \: \: or \: \: - \: \dfrac{ 1 }{ \sqrt{2} }$

Now, using first derivative test

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Interval & \bf Sign \: of \: f'(x) \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf x < - \dfrac{1}{ \sqrt{2} } & \sf - ve \\ \\ \sf - \dfrac{1}{ \sqrt{2} } < x < \dfrac{1}{ \sqrt{2} } & \sf + ve\\ \\ \sf x > \dfrac{1}{ \sqrt{2} } & \sf - ve \end{array}} \\ \end{gathered}$

$\bf\implies \:f(x) \: is \: minimum \: at \: x \: = \: - \: \dfrac{1}{ \sqrt{2} } \\ \\ and \\ \\ \bf\implies \:f(x) \: is \: maximum \: at \: x \: = \: \dfrac{1}{ \sqrt{2} }$

So,

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Point & \bf Value \: of \: f(x) \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf - \dfrac{1}{ \sqrt{2} } & \sf - \dfrac{1}{2} \\ \\ \sf \dfrac{1}{ \sqrt{2} } & \sf \dfrac{1}{2} \end{array}} \\ \end{gathered}$

Hence,

$\rm \implies\:\boxed{ \tt{ \: Range \: of \: f(x) = \bigg[ - \dfrac{1}{2}, \: \dfrac{1}{2} \bigg] \: }}$

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More to Know :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

$\large\underline{\text{}}$$\large\underline{\text{Solution}}$

$\red{\bigstar}$The domain of the function.

For the surd to be defined,

$\implies1-x^{2}\geq0$

$\implies x^{2}-1\leq0$

$\implies(x+1)(x-1)\leq0$

$\implies -1\leq x\leq1$

$\red{\bigstar}$Characteristic of the graph.

However

$\implies f(x)=x\sqrt{1-x^{2}}$

$\implies f(-x)=-x\sqrt{1-x^{2}}$

Hence

$\implies f(x)=-f(-x)$

Thereby the graph $f(x)=x\sqrt{1-x^{2}}$ is symmetric against the origin.

$\red{\bigstar}$Condition for the A.M-G.M inequality.

The A.M-G.M inequality can be used when either A.M or G.M is a constant. However, it can be applied when the numbers are positive or 0.

$\red{\bigstar}$Condition to satisfy the equality.

The inequality shows an equal sign when two numbers are equal.

$\red{\bigstar}$(Application) Condition for the A.M-G.M inequality.

Where $0\leq x\leq1$, consider the two numbers which are positive or 0.

$\implies x\geq0$

$\implies\sqrt{1-x^{2}}\geq0$

Squaring both

$\implies x^{2}\geq0$

$\implies1-x^{2}\geq0$

It is seen that the A.M of two numbers is $\dfrac{1}{2}$. Since the condition for A.M exists, we can use A.M-G.M inequality.

$\implies \dfrac{x^{2}+(1-x^{2})}{2}\geq\sqrt{x^{2}(1-x^{2})}$

$\implies \dfrac{1}{2}\geq\sqrt{x^{2}(1-x^{2})}$

$\implies\sqrt{x^{2}(1-x^{2})}\leq\dfrac{1}{2}$

That is

$\implies0\leq x\sqrt{1-x^{2}}\leq\dfrac{1}{2}\text{ for }0\leq x\leq1$

$\red{\bigstar}$(Application) Characteristic of the graph.

However, the graph is symmetric against the origin.

$\implies-\dfrac{1}{2}\leq x\sqrt{1-x^{2}}\leq\dfrac{1}{2}$

That is

$\implies-\dfrac{1}{2}\leq f(x)\leq\dfrac{1}{2}$

$\red{\bigstar}$(Application) Condition for the A.M-G.M inequality.

It is maximum when $x=\sqrt{1-x^{2}},x\geq0$.

According to the condition of equality, maximum and minimum occur when $x=\dfrac{1}{\sqrt{2}}$ and $x=-\dfrac{1}{\sqrt{2}}$ respectively.

$\large\underline{\text{Conclusion}}$

The range of the function is $f(x)\in[-\dfrac{1}{2},\dfrac{1}{2}]$ and the domain is $x\in[-1,1]$.