# Show that there is a number c, with 0\leq c\leq 1, such that f'(c)=0 for f(x)=x-\cos x.

## Question:

Show that there is a number {eq}c {/eq}, with {eq}0\leq c\leq 1 {/eq}, such that {eq}f'(c)=0 {/eq} for {eq}f(x)=x-\cos x {/eq}.

## Rolle's Theorem

If we have a continuous and differentiable function, we can see if we can apply Rolle's Theorem on an interval. Rolle's Theorem states that if the function values at the endpoints of an interval are equal, the first derivative must be zero at least once in that interval.

## Answer and Explanation:

Our function, {eq}f(x)=x-\cos x {/eq}, is continuous and differentiable, so we can see if we satisfy the third hypothesis of Rolle's Theorem. To do so, we need to evaluate this function at the endpoints of this interval.

{eq}f(0)=(0)-\cos(0) = -1\\ f(1)=(1)-\cos(1) = 0.4597\\ {/eq}

This does not satisfy Rolle's Theorem. While this does not necessarily mean there isn't a value in this interval causing the first derivative is equal to zero, it does mean that such a value is not guaranteed. There could be a smaller interval that could satisfy the required hypothesis.

We could attempt to find a number anyway if we wish, but we will be unsuccessful:

{eq}f'(x) = 1 + \sin x = 0\\ \sin x = -1\\ x = \frac{3 \pi}{2} {/eq}

However, {eq}\frac{3 \pi}{2} {/eq} is not within our interval. Thus, the statement of the problem is incorrect -- there is not a number in the interval where the first derivative is equal to zero.

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