show that the function f(x)= sqrt (x^2 +1) satisfies the 2 hypotheses of the Mean Value Theorem
f(x) = sqrt(x^2 + 1) ; [(0, sqrt(8)]
Okay, so for the Mean Value Theorem, two things have to be true: f(x) has to be continuous on the interval [0, sqrt(8)] and f(x) has to be differentiable on the interval (0, sqrt(8)).
First find where sqrt(x^2 + 1) is continous on. We know that for square roots Read More: ...