# Rolle's Theorem

2021-10-03

Verify that the hypotheses of Rolle’s Theorem are satisfied for f(x) = $$1\over6$$x - $$\sqrt {x}$$ on the interval [0,36], and find the value of c in this interval that satisfies the conclusion of the theorem.

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Rolle's Theorem: If a real-valued function f(x) is continuous on a closed interval [a,b], differentiable on the open interval (a, b), and f(a) = f(b), then there is some real number c in the open interval (a, b) such that f (0) = 0

$$f(x)=\frac{1}{6}x-\sqrt x$$ $$[0,36]$$

$$→ f(x)$$ is continuous on $$[0,36]$$
$$→ f(x)$$ is differentiable on $$(0,36)$$
$$→ f(0) = f(36) = 0$$

Also $$f '(x)= \frac{1}{6}-\frac{1}{2\sqrt x}$$

$$f'(0)=\frac{1}{6}-\frac{1}{2\sqrt x}$$

$$f'(36)=\frac{1}{6}-\frac{1}{2\sqrt 36}=\frac{1}{12}$$

$$⇒$$ all conditions of Rolle's Theorem are satisfied

So answer $$C=\frac{1}{12}$$