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Mathematics Question on Application of derivatives

If f(x)=x2−1x2+1,x∈Rf\left(x\right) = \frac{x^{2} -1}{x^{2} +1} ,x\in R then the minimum value of ff is

A

0

B

44656

C

44625

D

-1

Answer

-1

Explanation

Solution

We have, f(x)=x2−1x2+1f\left(x\right) = \frac{x^{2} -1}{x^{2} +1}
Maximum or minimum value point is given by f′(x)=0f'(x) = 0
f′(x)=(x2+1)2x−(x2−1)2x(x2+1)2=0f'\left(x\right) =\frac{\left(x^{2}+1\right)2x-\left(x^{2}-1\right)2x}{\left(x^{2}+1\right)^{2}} =0
⇒2x(x2+1−x2+1)=0\Rightarrow2x\left(x^{2}+1-x^{2}+1\right)=0
⇒4x=0⇒x=0\Rightarrow 4x =0 \Rightarrow x=0
Now, f′′(x)=ddx(2x3+2x−2x3+2x(x2+1)2)f''\left(x\right) = \frac{d}{dx} \left(\frac{2x^{3} +2x -2x^{3} +2x}{\left(x^{2} +1\right)^{2}} \right)
=ddx(4x(x2+1)2)=\frac{d}{dx}\left(\frac{4x}{\left(x^{2}+1\right)^{2}}\right)
f′′(x)=(x2+1)24−4x×2(x2+1)×2x(x2+1)4f''\left(x\right)= \frac{\left(x^{2}+1\right)^{2} 4 -4x\times2\left(x^{2}+1\right)\times 2x}{\left(x^{2}+1\right)^{4}}
=4(x2+1)2−16x2(x2+1)(x2+1)4= \frac{4\left(x^{2}+1\right)^{2}-16x^{2}\left(x^{2}+1\right)}{\left(x^{2}+1\right)^{4}}
=4(x2+1)(x2+1−4x2)(x2+1)4=4(−3x2+1)(x2+1)3= \frac{4\left(x^{2}+1\right)\left(x^{2}+1 -4x^{2}\right)}{\left(x^{2}+1\right)^{4}} =\frac{4\left(-3x^{2}+1\right)}{\left(x^{2}+1\right)^{3}}
Now f′′(0)=41>0f''(0) = \frac{4}{1} > 0
∴  x=0\therefore \:\: x = 0 is a minimum point and minimum value of ff is given by f(0)=0−10+1=−1 f(0) = \frac{0-1}{0+1} = - 1