Hi, please could someone help me with the following

f(x) = square root (x), x>0

Apply the Mean Value Theorem to f(x) on the interval [100,102] to prove that

111/11 < sqrt (102) < 101/10

I've proved that sqrt (102) < 101/10 by saying that there exists c such that the gradient at c = ( f(102) - f(100) ) / (102-100), then by saying that the gradient at c is less than the gradient at x=100

Annoyingly, if i try saying the gradient at c is greater than the gradient at x=102 then I end up with exactly the same inequality. How do i prove the left hand inequality, 111/11 < sqrt (102)?

Thanks.