Respuesta :
Answer:
See the explanation.
Step-by-step explanation:
Given function [tex]f(x)=x^3-x^2-6x+2[/tex]
And the interval [tex][0,3][/tex]
According to Rolle's Theorem
Let [tex]f(x)[/tex] be differentiable on the open interval [tex](a,b)[/tex] and continuous on the closed interval [tex][a,b][/tex]. Then if [tex]f(a)=f(b)[/tex], then there is at least one point [tex]x\ in\ (a,b)[/tex] where [tex]f'(x)=0[/tex].
So,
[tex]f(0)=0^3-0^2-6\times0+2\\\\f(0)=2\\\\Similarly\\\\f(3)=3^3-3^2-6\times3+2\\\\f(3)=27-9-18+2\\\\f(3)=2\\\\We\ can\ see\ f(0)=f(3)\\\\Now,\\\\f'(x)=\frac{d}{dx}(x^3-x^2-6x+2)\\\\f'(x)=3x^2-2x-6\\\\put\ f'(x)=0\\\\3x^2-2x-6=0\\[/tex]
We will find the value of [tex]x[/tex] for which [tex]f'(x)[/tex] became zero.
[tex]If\ ax^2+bx+c=0\\\\Then,\ x_{1}=\frac{-b+\sqrt{b^2-4ac}}{2a}\\ \\And\ x_{2}=\frac{-b-\sqrt{b^2-4ac}}{2a}\\\\f'(x)=3x^2-2x+6=0\\a=3,\ b=-2,\ c=6\\\\\ x_{1}=\frac{-(-2)+\sqrt{(-2)^2-4\times3\times(-6)}}{2\times3}\\\\x_{1}=\frac{2+\sqrt{76}}{6}=\frac{2+2\sqrt{19}}{6}\\\\x_{1}=\frac{1+\sqrt{19}}{3}\\\\x_{2}=\frac{-(-2)-\sqrt{(-2)^2-4\times3\times(-6)}}{2\times3}\\\\x_{1}=\frac{2-\sqrt{76}}{6}=\frac{2-2\sqrt{19}}{6}\\\\x_{2}=\frac{1-\sqrt{19}}{3}[/tex]
We can see
[tex]x_{1}=\frac{1+\sqrt{19}}{3}=1.786\\\\and\ 1.786\ is\ in\ [0,3][/tex]
There is at least one point [tex]1.786\ in\ (0,3)[/tex] where [tex]f'(x)=0[/tex].
