A) a local maxima at \[x=1\]and a local minima at \[x=-1\]
B) a local minima at \[x=1\] and a local maxima at \[x=-1\]
C) absolute maxima at \[x=1\] and absolute minima at \[x=-1\]
D) Absolute minima at \[x=1\] and absolute maxima at \[x=-1\]
Correct Answer: B
Solution :
Given that, \[f(x)=x+\frac{1}{x}\] \[f'(x)=1-\frac{1}{{{x}^{2}}}\] For local maxima or local minima, put \[f'(x)=0\] \[\Rightarrow \] \[1-\frac{1}{{{x}^{2}}}=0\] \[\Rightarrow \] \[x=\pm 1\] Now, \[f''(x)=\frac{2}{{{x}^{3}}}\] At \[x=1,\] \[f''(x)=\frac{2}{{{1}^{3}}}\] \[=2>0,\] local minima At \[x=-1,\] \[f''(x)=\frac{2}{{{(-1)}^{3}}}\] \[=-2<0,\] local maxima
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