Question

f(x)=x³-6x²+11x-6 x ∈ [1,3],Verify Rolle's theorem

Answer 1

Answer:

f(x)=x³-6x²+11x-6 x ∈ [1,3],Verify Rolle's theorem

$f(x)=x^3-6x^2+11x-6$

$f'(x)=3x^2-12x+11$

$\text{since f(x) is a polynomial, }$

$\text{f(x) is continuous on [1,3]}$

$\text{f(x) is differentiable in (1,3)}$

$\text{Also, f(0)=0-0+0-6=-6}$

$\text{f(3)=27-54+33-6=0}$

$\implies\text{f(0)=f(3)}$

$\therefore\text{Conditions of rolle's theorem are satisfied}$

$\text{Now, f'(c)=0}$

$\implies\:3c^2-12c+11=0$

$\implies\:3c^2-12c+11=0$

$\text{using quadratic formula }$

$c=\frac{12\pm\sqrt{144-4(3)(11)}}{6}$

$c=\frac{12\pm\sqrt{144-132}}{6}$

$c=\frac{12\pm\sqrt{12}}{6}$

$c=\frac{12\pm2\sqrt{3}}{6}$

$c=\frac{6\pm\sqrt{3}}{3}$

$c=2\pm\frac{\sqrt{3}}{3}$

$c=2\pm\frac{1}{\sqrt{3}}$

$\text{Clearly }2\pm\frac{1}{\sqrt{3}}\in(1,3)$

$\text{Hence rolle's theorem verified.}$