Let f(x) = x2 + (1/x2) and g(x) = x - (1/x) , x ∈ R - - 1, 0, 1 . If h(x) = (f(x) /g(x)), then the local minimum value of h(x) is

Question

Let f(x) = x2 + (1/x2) and g(x) = x - (1/x) , x ∈ R - - 1, 0, 1 . If h(x) = (f(x) /g(x)), then the local minimum value of h(x) is (A) 3 (B) -3 (C) - 2 √2 (D)  2 √2

Tardigrade Admin · Accepted Answer

h(x)=(x2+(1/x2)/x-1 / x) =(x-1 / x)+(2/(x-1 / x)) x-(1/x)>0,(x-1 / x)+(2/(x-1 / x)) ∈(2 √2, ∞] x-(1/x)<0,(x-1 / x)+(2/(x-1 / x)) ∈(-∞,-2 √2] Local minimum is 2 √2

Q. Let $f (x) = x^{2} + \frac{1}{x ^{2}}$ and $g (x) = x - \frac{1}{x}, x \in R - {- 1, 0, 1}$ . If $h (x) = \frac{f ( x )}{g ( x )}$ , then the local minimum value of $h (x)$ is

3

-3

$- 22$

$22$

$h (x) = \frac{x ^{2} + \frac{1}{x ^{2}}}{x - 1/ x}$
$= (x - 1/ x) + \frac{2}{( x - 1/ x )}$
$x - \frac{1}{x} > 0, (x - 1/ x) + \frac{2}{( x - 1/ x )} \in (22, \infty]$
$x - \frac{1}{x} < 0, (x - 1/ x) + \frac{2}{( x - 1/ x )} \in (- \infty, - 22]$
Local minimum is $22$

Q. Let f(x)=x2+x21​ and g(x)=x−x1​,x∈R−{−1,0,1}. If h(x)=g(x)f(x)​, then the local minimum value of h(x) is

3

-3

−22​

22​

h(x)=x−1/xx2+x21​​ =(x−1/x)+(x−1/x)2​ x−x1​>0,(x−1/x)+(x−1/x)2​∈(22​,∞] x−x1​<0,(x−1/x)+(x−1/x)2​∈(−∞,−22​] Local minimum is 22​

Q. Let $f (x) = x^{2} + \frac{1}{x ^{2}}$ and $g (x) = x - \frac{1}{x}, x \in R - {- 1, 0, 1}$ . If $h (x) = \frac{f ( x )}{g ( x )}$ , then the local minimum value of $h (x)$ is

$- 22$

$22$

$h (x) = \frac{x ^{2} + \frac{1}{x ^{2}}}{x - 1/ x}$
$= (x - 1/ x) + \frac{2}{( x - 1/ x )}$
$x - \frac{1}{x} > 0, (x - 1/ x) + \frac{2}{( x - 1/ x )} \in (22, \infty]$
$x - \frac{1}{x} < 0, (x - 1/ x) + \frac{2}{( x - 1/ x )} \in (- \infty, - 22]$
Local minimum is $22$