Question

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

• A. 1
• B. $\sqrt{3}$
• C. 2
• D. $\sqrt{2}$

Answer 1

The correct option is B $\sqrt{3}$

Since, the Lagrange's Mean Value Theorem holds for the given function $f\left(x\right)=x+\frac{1}{x},x\in \left(1,3\right]$. Therefore, there exists a value c in [1, 3] such that $f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}\Rightarrow f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}\Rightarrow 1-\frac{1}{c^2}=\frac{3+\frac{1}{3}-\left(1+1\right)}{2}\left(\because f\text{'}\left(x\right)=1-\frac{1}{x^2}\right]\Rightarrow 1-\frac{1}{c^2}=\frac{\frac{4}{3}}{2}=\frac{2}{3}\Rightarrow \frac{1}{c^2}=\frac{1}{3}\Rightarrow c=\sqrt{3}.$