A) 2
B) 1
C) -1
D) -2
Correct Answer: B
Solution :
Let \[f(x)=x+\frac{1}{x}\] \[f'(x)=1-\frac{1}{{{x}^{2}}}\] Put \[\,f'(x)=0\] for maxima and minima. \[\Rightarrow \] \[1-\frac{1}{{{x}^{2}}}=0\] \[\Rightarrow \] \[x=\pm 1\] \[f''(x)\frac{2}{{{x}^{3}}}\] At \[x=1,\,f''(x)=+ve\] (minima) and at \[x=-1\], \[f''(x)=-ve\] (maxima)You need to login to perform this action.
You will be redirected in
3 sec