Prove that $$\sqrt{1+x} < 1+\frac{1}{2}x \enspace $$ for $x > 0$ by using the mean value theorem:
we have that if the function is continuous at some point the secant line is parallel to the tangent line $$f'(x) = \frac{f(b)-f(a)}{b-a}$$
Lets take $f(x) =\sqrt{1+x} $ $$f'(x) = \frac{1}{2\sqrt{1+x}}$$ now $g(x) = 1+ \frac{1}{2}x $ $$g'(x) = \frac{1}{2}$$
How can I continue this proof?