Verify that each function satisfies the hypothesis of the Mean Value theorem on the interval [a,b] and find all the numbers c that satisfy the conclusion.. ..of the Mean Value Theorem. f(x) = 1 - (sqrt)x-1 on [2,10]
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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):
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