Question

Asked Apr 26, 2019

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For the function f(x)= (x^2)-(4x)+6 on the interval [0,4]

What is the value of f(x) on [0,4]?

What is the value of f(x) on (0,4)?

If f(0)= f(4)= ?

By Rolle's Theroem, f '(c)=0

what does c=?

Step 1

The given function is *f* (*x*) = *x*^{2 }– 4*x *+ 6 with the interval [0,4].

It is known that, the Rolle’s theorem states, If a real valued function *f* is continuous on a proper closed interval [*a*, *b*], differentiable on the open interval (*a*, *b*), and *f* (*a*) = *f* (*b*), then there exist at least one *c* in the open interval (*a*, *b*) such that *f *’ (*c*) = 0.

Step 2

By the above theorem, the function *f* (*x*) = *x*^{2 }– 4*x *+ 6 continuous on closed interval [0,4], and differentiable on the open interval (0,4).

Find the value of *f* (0) as follows.

*f* (0) = (0)^{2 }– 4(0) + 6

= 0 – 0 + 6

=6

Find the value of *f* (4) as follows.

*f* (4) = (4)^{2 }– 4(4) + 6

= 16 – 16 + 6

=0 + 6

= 6

Thus, the both values *f* (0) and *f* (4) is 6.

Step 3

It is given that, by Rolle’s theorem ...

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