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:

Name: Video solution 1: 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:
Uploaded: 2021-01-01T18:41:02.000Z
Duration: 2 min
Description: Video solution 1: 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:

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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:

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Elizabeth Discussed

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:

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Question Text	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:
Answer Type	Video solution: 1
Upvotes	150

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:

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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: