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=? 

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Expert Answer

Step 1

The given function is f (x) = x2 – 4x + 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) = x2 – 4x + 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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Calculus

Functions

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