For the function $f(x)=x+\frac{1}{x},x\in [1,3]$, the value of $c$ for Lagrange's Mean Value Theorem is:

According to Mean Value Theorem, there exists $c\in (1,3)$ such that:
$$f^{\prime }(c)=\frac{f(3)-f(1)}{3-1}$$
Find $f(x)$ values:$f(1)=1+\frac{1}{1}=2$$f(3)=3+\frac{1}{3}=\frac{10}{3}$ Find $f^{\prime }(x)$:$f^{\prime }(x)=1-\frac{1}{x^2}$ Substitute into the MVT formula:
$$1-\frac{1}{c^2}=\frac{\frac{10}{3}-2}{2}$$
$$1-\frac{1}{c^2}=\frac{\frac{4}{3}}{2}=\frac{2}{3}$$
$$\frac{1}{c^2}=1-\frac{2}{3}=\frac{1}{3}$$
$$c^2=3⟹c=\sqrt{3}$$
Since $\sqrt{3}\approx 1.732$ lies in $(1,3)$, $c=\sqrt{3}$.