Given `f(x)=1-sqrt(x-1)` on the interval [2,10]

The function is continuous on [2,10] and differentiable on (2,10) so we can use the mean valuetheorem.

The MVT states that there exists `c in (a,b)` such that `f'(c)=(f(b)-f(a))/(b-a)` or the value of the derivative at some c is equal to the slope of...

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Given `f(x)=1-sqrt(x-1)` on the interval [2,10]

The function is continuous on [2,10] and differentiable on (2,10) so we can use the mean valuetheorem.

The MVT states that there exists `c in (a,b)` such that `f'(c)=(f(b)-f(a))/(b-a)` or the value of the derivative at some c is equal to the slope of the secant line through the endpoints of the interval.

Now `(f(10)-f(2))/(10-2)=(-2-0)/8=-1/4` ; by the MVT there exists `c in (2,10)` such that `f'(c)=-1/4` .

`f'(x)=-1/(2sqrt(x-1))`

`-1/(2sqrt(x-1))=-1/4`

`1/sqrt(x-1)=1/2`

`sqrt(x-1)=2`

`x-1=4` ==>x=5

**So the required c is 5.**

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The graph of the function; the secant line, and the tangent line through (5,-1):