verify Rolle's theorem f(x)=x^2_4x+3
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Let you assume that |9−x2||9−x2| is not differentiable on −3−3 and 33, so the derivatives cannot be zero at that point, as they do not exist at all. You must have made a mistake somewhere.
In your problem particularly, as f(x)=9−x2f(x)=9−x2 we have f′(x)=−2xf′(x)=−2x, as 9−x2≥09−x2≥0 on [−3,3][−3,3], the point where the derivative is 00 is 00, and only 00.
Also, note that Rolle's Theorem says that the point where the derivatives are zero are all in [a,b][a,b] where f(a)=f(b)f(a)=f(b). It merely says that there exists such a point.
For example, take f(x)=0f(x)=0 on [1,3][1,3]. There exists a value where f′(x)=0f′(x)=0 on (1,3)(1,3), but f′(1)=f′(3)=0f′(1)=f′(3)=0 as well. So values where f′(x)=0f′(x)=0 need not always exist in the interval (a,b)(a,b).
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In your problem particularly, as f(x)=9−x2f(x)=9−x2 we have f′(x)=−2xf′(x)=−2x, as 9−x2≥09−x2≥0 on [−3,3][−3,3], the point where the derivative is 00 is 00, and only 00.
Also, note that Rolle's Theorem says that the point where the derivatives are zero are all in [a,b][a,b] where f(a)=f(b)f(a)=f(b). It merely says that there exists such a point.
For example, take f(x)=0f(x)=0 on [1,3][1,3]. There exists a value where f′(x)=0f′(x)=0 on (1,3)(1,3), but f′(1)=f′(3)=0f′(1)=f′(3)=0 as well. So values where f′(x)=0f′(x)=0 need not always exist in the interval (a,b)(a,b).
PLEASE MARK MY ANSWER AS A BRAINLIEST ANSWER.
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