Comment
Share
Q)

# Verify Lagrange's theorem for the following function;$f(x)=x^{3}-5x^{2}-3x\;[1,3]$

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

Comment
A)
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 $f'(c)=0$
$f(x)=x^3-5x^2-3x \qquad[1,3]$
Step 1:
$f(x)$ is a polynomial function .
It is continuous in $[1,3]$ and differentiable in $(1,3)$
The conditions for Lagranges Theorem are satisfied.
Step 2:
To find $c \in (1,3)$
Such that $f'(c)=\large\frac{f(3)-f(1)}{3-1}$
$f'{x}=3x^2-10 x -3$
$f'(c)=3c^2-10c-3$
$\large\frac{f(3)-f{(1)}}{3-1}=\large\frac{27-45-9-(1-5-3)}{2}=\frac{-27+7}{2}$
$3c^2-10 c-3=-10$
$3c^2-10 c+7=0 \qquad c=\large\frac{10 \pm \sqrt {100-84}}{6}=\frac{10 \pm 4}{6}=\frac{14}{6},$$1 \large\frac{14}{6} =\frac{7}{3} \in$$(1,3)$ is the required 'c'
Lagrange's Theorem is verified