Browse Questions

Verify Lagrange's law of mean for the following function;$f(x)=1-x^{2} \quad [0,3]$.

Note: This is part 1st of a 5 part question, split as 5 separate questions here.

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 $f(x)=1-x^2\;[0,3]$
Step 1:
$f(x)$ is a polynomial function
So it is conditions in $[0,3]$ and differentiable in $(0,3)$ the condition's for Lagrange's
Step 2:
To find $c \in (0,3)$where $f'(c)=\large\frac{f(b)-f(a)}{b-a}$
$f'(x)=-2x$
$f'(c)=-2c$
$-2c=\large\frac{f(3)-f(0)}{3-0}$
$-2c=\large\frac{(1-9)-(1)}{3}$$=-3 =>c=\large\frac{3}{2} \large\frac{3}{2}$$ \in (0,3)$. Lagrange's Theorem is verified