I need help finding the relative extrema for : f(x) = square root of x^2+1
, by using the first and second derivative tests
$\displaystyle f(x)=\sqrt{x^2 + 1}$
then,
$\displaystyle f'(x)=\frac{x}{\sqrt{x^2 + 1}}$
then, get the critical point,
$\displaystyle f'(x)=x=0$
so,
x=0
then,
$\displaystyle f''(x)=\frac{\sqrt{x^2 + 1} - x\frac{x}{\sqrt{x^2 + 1}}}{x^2 + 1}$
if you simplify this, you will get
$\displaystyle f''(x)= \frac{1}{(\sqrt{x^2 + 1})(x^2 + 1)}$
which is always positive, therefore the function assumes a minimum point at x=0. so by first and second derivative test, the relative minimum point of
$\displaystyle f(x)=\sqrt{x^2 + 1}$
is at (0,1)..