Question

Verify rolles theorem for f(x) =x^3-x^2-6x+2 in [0,3]​

Answer 1

See the explanation.

Explanation:

Given function
`f(x)=x^3-x^2-6x+2`

And the interval
`[0,3]`

According to Rolle's Theorem

Let
`f(x)` be differentiable on the open interval
`(a,b)` and continuous on the closed interval
`[a,b]`. Then if
`f(a)=f(b)`, then there is at least one point
`x\ in\ (a,b)` where
`f'(x)=0`.

So,

`f(0)=0^3-0^2-6*0+2\\\\f(0)=2\\\\Similarly\\\\f(3)=3^3-3^2-6*3+2\\\\f(3)=27-9-18+2\\\\f(3)=2\\\\We\ can\ see\ f(0)=f(3)\\\\Now,\\\\f'(x)=(d)/(dx)(x^3-x^2-6x+2)\\\\f'(x)=3x^2-2x-6\\\\put\ f'(x)=0\\\\3x^2-2x-6=0\\`

We will find the value of
`x` for which
`f'(x)` became zero.

`If\ ax^2+bx+c=0\\\\Then,\ x_(1)=(-b+√(b^2-4ac))/(2a)\\ \\And\ x_(2)=(-b-√(b^2-4ac))/(2a)\\\\f'(x)=3x^2-2x+6=0\\a=3,\ b=-2,\ c=6\\\\\ x_(1)=(-(-2)+√((-2)^2-4*3*(-6)))/(2*3)\\\\x_(1)=(2+√(76))/(6)=(2+2√(19))/(6)\\\\x_(1)=(1+√(19))/(3)\\\\x_(2)=(-(-2)-√((-2)^2-4*3*(-6)))/(2*3)\\\\x_(1)=(2-√(76))/(6)=(2-2√(19))/(6)\\\\x_(2)=(1-√(19))/(3)`

We can see

`x_(1)=(1+√(19))/(3)=1.786\\\\and\ 1.786\ is\ in\ [0,3]`

There is at least one point
`1.786\ in\ (0,3)` where
`f'(x)=0`.