MHB Is the given function differentiable at the endpoints on the interval [0,5]?

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The function f(x) = √(x(5-x)) is continuous on the interval [0,5] but not differentiable at the endpoints due to vertical slopes. The Mean Value Theorem (MVT) requires the function to be continuous on the closed interval and differentiable on the open interval, which means differentiability at endpoints is not necessary for MVT to apply. The discussion clarifies that the MVT can still hold even if the derivative is undefined at the endpoints. Understanding the MVT involves recognizing that the gradient of any chord corresponds to the gradient of the function at some point within the interval. Therefore, the function satisfies the conditions of the MVT despite the endpoints lacking differentiability.
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Does the function satisfy the hypothesis of the mean value theorm
On the given interval, give reasons for your answer

$$f\left(x\right)=\sqrt{x\left(5-x\right)},\ \ \left[0,5\right]$$

The graph of this is the top half of a circle

My answer was
It is continuous at all points on the interval but not differential at the endpoints due to vertical slopes.

This answer was not correct.?

MVT $$f'\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$$
 
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karush said:
Does the function satisfy the hypothesis of the mean value theorm
On the given interval, give reasons for your answer

$$f\left(x\right)=\sqrt{x\left(5-x\right)},\ \ \left[0,5\right]$$

The graph of this is the top half of a circle

My answer was
It is continuous at all points on the interval but not differential at the endpoints due to vertical slopes.

This answer was not correct.?

MVT $$f'\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$$

What is your hypothesis about the MVT?
 
Not sure, the book didn't address where $f'$ is undefined
 
karush said:
Not sure, the book didn't address where $f'$ is undefined

I don't think you understand the mean value theorem very well. It states that if your function is continuous everywhere and smooth at all points except the endpoints, then the gradient of any chord on that function is the same as the gradient of the function at some point in between. So your derivative doesn't need to be defined at the endpoints...
 
OK, thanks.
 
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