# F(x) is a function on interval [1, 5] &suppose that F′(x) exists for all x ∈ (1, 5). If F(1)=2 and F(5)=2,why is F(x) a critical number c in the interval [1,5].

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### 1 Answer

F is differentiable on the open interval (1,5) and continuous on the closed interval [1,5] and F(1)=F(5)=2.

This function fits the requirements for Rolle's theorem: If F is continuous on the closed interval [a,b] and differentiable on the open interval (a,b) and F(a)=F(b) then there exists `cin(a,b)` such that `f'(c)=0` and thus c is a critical number for F.

If you are being asked to prove this instance of Rolle's theorem:

Let f(a)=2=f(b):

Case 1: If f(x)=2 for all `x in (a,b)` then the derivative at any x is zero.

Case 2: Suppose f(x)>2 for some `x in (a,b)` . By the Extreme Value Theorem there is a maximum at some `c in [a,b]` ; since `f(c)>2` it does not occur at an endpoint. Thus f(x) has a maximum on the open interval and f(c) is a relative maximum which can only occur if `f'(x)=0` or fails to exist. Since the function is differentiable, `f'(c)=0` .

Case 3: Argue analogously from case 2 if f(x)<2 for some `x in (a,b)`