Choose the correct answers from the given four options: For the function $f(x)=x+\frac{1}{x}$, $x \in [1,3]$…
Choose the correct answers from the given four options:
For the function $f(x)=x+\frac{1}{x}$, $x \in [1,3]$, the value of $c$ for mean value theorem is:
$\sqrt{3}$
$\sqrt{2}$
$1$
$2$
Solution
The Mean Value Theorem states that if a function $f(x)$ is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, then there exists at least one number $c$ in the interval $(a, b)$ such that the derivative of the function at that point is equal to the average rate of change of the function over the interval.
Given the function $f(x) = x + \frac{1}{x}$, $x \in [1,3]$, we first find the derivative of the function.
$
f'(x) = 1 - \frac{1}{x^2}
$
According to the Mean Value Theorem, there exists a $c$ in the interval $(1, 3)$ such that
$
f'(c) = \frac{f(3) - f(1)}{3 - 1}
$
Substituting $f(x)$ and $f'(x)$ into the equation, we get
$
1 - \frac{1}{c^2} = \frac{(3 + \frac{1}{3}) - (1 + 1)}{3 - 1}
$
Solving this equation gives
$
1 - \frac{1}{c^2} = \frac{1}{2}
$
$
\frac{1}{c^2} = \frac{1}{2}
$
$
c^2 = 2
$
$
c = \sqrt{2}
$
But $\sqrt{2}$ is not in the given options.
The reason for this discrepancy is that we only considered the positive root of $c^2 = 2$. However, since $c$ is in the interval $(1, 3)$, we should consider the negative root as well.
So,
$
c = -\sqrt{2}
$
But $-\sqrt{2}$ is not in the interval $(1, 3)$.
This means that there is no value of $c$ in the interval $(1, 3)$ that satisfies the Mean Value Theorem for the given function and the given interval.
Therefore, none of the options A) $\sqrt{3}$, B) $\sqrt{2}$, C) $1$, or D) $2$ are correct.