Question

The real number $x$ when added to its inverse gives the minimum value of the sum at $x$ equal to

• A. $1$
• B. $-1$
• C. $-2$
• D. $2$

Answer 1

Answer: (C)

$-2$

Let the number be x.
Hence
$f\left(x\right)=x+\frac{1}{x}$
$\frac{d\left(f\right(x\left)\right)}{dx}=1-\frac{1}{x^2}$
for critical points, $f^{\prime }\left(x\right)=0$
Or
$1-\frac{1}{x^2}=0$
$x^2=1$
$x=\pm 1$
Now
$f\left(1\right)=2$ and $f(-1)=-2$
Thus $f\left(x\right)$ attains a minimum value of 2 at $x=-1$.
Hence the minimum value of the sum of a number and its reciprocal is -2.