# Verify Lagranges theorem for the following function;$f(x)=\large\frac{1}{x}$$\quad[1,2] Note: This is part 2nd of a 5 part question, split as 5 separate questions here. ## 1 Answer Toolbox: • Lagrange's Mean Value Theorem : • Let f(x) be a real valued function that satisfies the following conditions. • (i) f(x) is continuous on the closed interval [a,b] • (ii) f(x) is differentiable in the open interval (a,b) • (iii) f(a)=f(b) • Then there exists atleast one value c \in (a,b) such that f'(c)=0 Given f(x)=\large\frac{1}{x}$$\;[1,2]$
Step 1:
The function is continuous in $[1,2]$ and differentiable in $(1,2)$. The conditions for lagranges theorem are satisfied .
Step 2:
To find $c \in (1,2)$ where $f'(c)=\large\frac{f(2)-f(1)}{2-1}$
$f'(x)=\large\frac{-1}{x^2}$
$f'(c)=\large\frac{-1}{c^2}$
$\large\frac{f(2)-f(1)}{2-1}=\frac{\Large\frac{1}{2}-1}{2-1}=\frac{-1}{2}$
$\large\frac{-1}{c^2}=\large\frac{-1}{2}$$=>c=\pm \sqrt 2$
Of these values $\sqrt 2 \in (1,2)$ the theorem is verified