mean value theorem prove sq root 9.1 is less than or equal to 3+1/60
let f(x) = x^2-9.1
f(3) = = -.1
f(3+1/60) = 9 + 6/60 + 1/3600 - 9.1 = 1/3600
since f is continuous, f assumes all values between -.1 and 1/3600, which includes 0.
f(3) = = -.1
f(3+1/60) = 9 + 6/60 + 1/3600 - 9.1 = 1/3600
since f is continuous, f assumes all values between -.1 and 1/3600, which includes 0.