Consider the function f(x)=10sqrt(x)+8on the interval [3,9]. Find the average or mean slope of the function on this interval.

By the Mean Value Theorem, we know there exists a c in the open interval (3,9) such that f'(c) is equal to this mean slope. For this problem, there is only one c that works. Find it.

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You need to use mean value theorem over interval [3,9] such that:

`f'(c) = (f(b)-f(a))/(b-a)`

`f'(c) = (f(9)-f(3))/(9-3)`

You need to evaluate f(9) and f(3) such that:

`f(9) = 10sqrt9 + 8`

`f(9) = 38`

`f(3) = 10sqrt3 + 8`

`f'(c) = (38 - 10sqrt3 - 8)/6 =gt f'(c) = (30 - 10sqrt3)/6`

You need to find derivative of the function such that:

`f'(x) = 10/(2sqrtx)`

`f'(x) = 5/sqrt x`

You need to evaluate f'(x) at x=c such that:

`f'(c) = 5/sqrt c`

Since `f'(c) = (30 - 10sqrt3)/6` , then `5/sqrt c = (30 - 10sqrt3)/6` such that:

`5/sqrt c = 10(3 - sqrt3)/6 =gt 1/sqrt c =2(3 - sqrt3)/6`

`1/sqrt c = (3 - sqrt3)/3 =gt 3sqrt c - sqrt (3c) = 3`

You need to factor out `sqrt c` such that:

`sqrt c(3 - sqrt 3) = 3 =gt sqrt c = 3/(3 - sqrt 3) `

`sqrt c = 3(3+sqrt3)/(9-3) =gt sqrt c = 3(3+sqrt3)/6`

`sqrt c = (3+sqrt3)/2 =gt c = ((3+sqrt3)^2)/4`

**Hence, evaluating the mean slope over interval (3,9) yields `f'(c) = 5(3-sqrt3)/3` .**

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