Question

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

• A. 1
• B. √3
• C. 2
• D. None of these

Answer 1

Answer: (B)

√3

Answer 2

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 value $c \in (a, b)$ such that:

$$f'(c) = \frac{f(b) - f(a)}{b - a}$$

Given the function $f(x) = x + \frac{1}{x}$ on the interval $[1, 3]$.

1. Check conditions for MVT:

   • The function $f(x) = x + \frac{1}{x}$ is continuous on $[1, 3]$ because $x \neq 0$ for any $x$ in this interval.
   • The function $f(x) = x + \frac{1}{x}$ is differentiable on $(1, 3)$ for the same reason.

   Thus, the conditions for the Mean Value Theorem are satisfied.

2. Calculate $f(a)$ and $f(b)$:

   Here, $a = 1$ and $b = 3$.

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

   $f(3) = 3 + \frac{1}{3} = \frac{9 + 1}{3} = \frac{10}{3}$

3. Calculate the slope of the secant line:

   $\frac{f(b) - f(a)}{b - a} = \frac{f(3) - f(1)}{3 - 1} = \frac{\frac{10}{3} - 2}{2}$

   $= \frac{\frac{10 - 6}{3}}{2} = \frac{\frac{4}{3}}{2} = \frac{4}{6} = \frac{2}{3}$

4. Calculate the derivative $f'(x)$:

   $f(x) = x + x^{-1}$

   $f'(x) = 1 - x^{-2} = 1 - \frac{1}{x^2}$

5. Set $f'(c)$ equal to the slope of the secant line and solve for $c$:

   $f'(c) = 1 - \frac{1}{c^2}$

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

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

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

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

   $c^2 = 3$

   $c = \pm\sqrt{3}$

6. Check if $c$ is in the interval $(1, 3)$:

   • For $c = \sqrt{3}$: We know that $1^2 = 1$ and $2^2 = 4$. Since $1 < 3 < 4$, it follows that $1 < \sqrt{3} < 2$. Therefore, $c = \sqrt{3}$ lies in the interval $(1, 3)$.
   • For $c = -\sqrt{3}$: This value is not in the interval $(1, 3)$.

Thus, the value of $c$ for the Mean Value Theorem is $\sqrt{3}$.