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Q.
The denominator of a fraction exceeds the square of the numerator by 16, then the least value of the fraction is
NTA AbhyasNTA Abhyas 2020Application of Derivatives
Solution:
Let the numerator be $x,$ then $f(x)=\frac{x}{x^{2}+16}$
On differentiating w.r.t. $x ,$ we get
$f^{\prime}(x)=\frac{\left(x^{2}+16\right) \cdot 1-x(2 x)}{\left(x^{2}+16\right)^{2}}$
$=\frac{x^{2}+16-2 x^{2}}{\left(x^{2}+16\right)^{2}}=\frac{16-x^{2}}{\left(x^{2}+16\right)^{2}} \ldots$ (i)
Putting $f ^{\prime}( x )=0$ for maxima or minima, we get,
$16-x^{2}=0 \Rightarrow x=4,-4$
Again, on differentiating w.r.t. $x ,$ we get
$f^{\prime \prime}(x)=\frac{\left(x^{2}+16\right)^{2}-(-2 x)-\left(16-x^{2}\right) 2\left(x^{2}+16\right) 2 x}{\left(x^{2}+16\right)^{4}}$
At $x=4, f^{\prime \prime}(x)<0$
$\therefore f ( x )$ is maximum at $x =4$
And at $x=-4, f^{\prime \prime}(x)>0 \Rightarrow f(x)$ is minimum
$\therefore $ Least value of $f(x)=\frac{-4}{16+16}=-\frac{1}{8}$