If M and m denote the local maximum and local minimum values of the function $\displaystyle f(x)=x+\frac{1}{x}(x \neq 0)$ respectively, find the value of $\displaystyle (M-m)$.

$$\begin{aligned} & f^{\prime}(x)=1-\frac{1}{x^{2}}=\frac{x^{2}-1}{x^{2}}=\frac{(x+1)(x-1)}{x^{2}} \\ & f^{\prime}(x)=0 \Rightarrow x=-1,1 \end{aligned} $\displaystyle \therefore-1$ is a point of local maximum The local maximum value $\displaystyle =f(-1)=-2=M$ f^{\prime \prime}(1)=2>0 $\displaystyle \therefore 1$ is point of local minimum The local minimum value $\displaystyle =f(1)=2=m$ M-m=-4 $$