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**HallsofIvy** You have the formula- just plug in the data for this problem.

The only "hard" part is finding the general derivative.

If $\displaystyle f(x)= \sqrt{x+ 1}= (x+ 1)^{1/2}$, then $\displaystyle f'(x)= (1/2)(x+ 1)^{-1/2}$, $\displaystyle f''(x)= -1/4(x+ 1)^{-3/2}$, $\displaystyle f'''(x)= 3/8(x+ 1)^{-5/2}$, $\displaystyle f^{(4)}= -15/16(x+1)^{-7/2}$...

Do you see the pattern? Obviously, there is an alternating + and -: $\displaystyle (-1)^{n+1}$ since this is + for n odd, - for n even.

The denominator of the fraction is a power of 2: $\displaystyle 2^n$.

The numerator is a little harder- it is a product of odd integers from 2n-1 down: 1*3*5*7... We can write that more compactly by filling in the missing evens: $\displaystyle \frac{1*2*3*4*5*6*7...}{2*4*6...}= \frac{n!}{2*4*6...}$. And we can write **that** denominator as $\displaystyle (2*1)*(2*2)(2*3)...= 2^n(1*2*3...)= 2^k k!$.

The fraction is $\displaystyle \frac{(2n-1)!}{(2^n)(2^{n-2}(n-2)!}$ and the exponent on x+ 1 is $\displaystyle \frac{-(2n-1)}{2}$.

That is, $\displaystyle f^(n)(x)= \frac{(-1)^{n+1}(2n-1)!}{2^{2n-2}(n-2)!}(x+ 1)^{-\frac{2n-1}{2}}$.

Using that, find a maximum on $\displaystyle f^{n+1}$ between 0 and x.