For the function f(x) = x + 1/x, x ∈ [1, 3] the value of c for mean value theorem is

The correct option is √3.

Step-by-step Explanation:

The Mean Value Theorem states that if a function $f\left(x\right)$ is continuous on a closed interval $[a,b]$ and differentiable on the open interval $(a,b)$, then there exists at least one value $c$ in the open interval $(a,b)$ such that:

$f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Here, the given function is $f\left(x\right)=x+\frac{1}{x}$ defined on the interval $[1,3]$. Thus, we have $a=1$ and $b=3$.

First, we calculate the values of the function at the endpoints of the interval:

$f\left(1\right)=1+\frac{1}{1}=2$

$f\left(3\right)=3+\frac{1}{3}=\frac{10}{3}$

Next, we compute the average rate of change on this interval:

$\frac{f\left(3\right)-f\left(1\right)}{3-1}=\frac{\frac{10}{3}-2}{2}=\frac{\frac{4}{3}}{2}=\frac{2}{3}$

Now, we find the derivative of the function $f\left(x\right)$ with respect to $x$:

$f^{\prime }\left(x\right)=1-\frac{1}{x^2}$

According to the theorem, we set the derivative evaluated at $c$ equal to the average rate of change:

$f^{\prime }\left(c\right)=1-\frac{1}{c^2}=\frac{2}{3}$

We solve for $c$ by rearranging the terms:

$\frac{1}{c^2}=1-\frac{2}{3}$

$\frac{1}{c^2}=\frac{1}{3}$

$c^2=3$

$c=\pm \sqrt{3}$

Since the value of $c$ must lie strictly within the open interval $(1,3)$, we select the positive root:

$c=\sqrt{3}\approx 1.732$

Since $\sqrt{3}$ lies in the interval $(1,3)$, this is the correct value of $c$ that satisfies the Mean Value Theorem.