Given a function f(x)= (x^3 + 1)/x  ,using calculus find the exact
values of the critical point (s) and find exact values of the point
(s) of inflection of f(x). I have started working through this using
the derivative product rule but got lost along the way- any help would
be much appreciated !

Hi pj4444!!!

I will start suggesting to see the following pages from MathWorld:

"Stationary Point":
http://mathworld.wolfram.com/StationaryPoint.html

"Critical Point":
http://mathworld.wolfram.com/CriticalPoint.html

"First Derivative Test":
http://mathworld.wolfram.com/FirstDerivativeTest.html

"Second Derivative Test":
http://mathworld.wolfram.com/SecondDerivativeTest.html


After you see these pages we can start with the exercise:

f(x)= (x^3 + 1)/x = (x^3 + 1) * 1/x = x^2 + 1/x (distributive
property),

So the domain of the function f is {(-oo , 0);(0, +oo)} i.e. all the
real numbers except zero, this is because zero is the only number for
that the function f is not defined.

At the first step I will calculate the first derivative function of f:
f´(x) = 2.x - 1/x^2    
(use the derivative of a sum is the sum of the derivatives property);
now we must find the critical points (points at which the derivative
of a function f(x) vanishes or f is not differentiable). We can see
that f is differentiable in all of its domain (remember that zero
isn´t be in its domain).

0 = 2.x - 1/x^2 ==> 2.x = 1/x^2 ==> 2.x^3 = 1 ==> x^3 = 1/2 ==> x =
(1/2)^1/3 ;

There is only one stationary (or critical) point is x0 = (1/2)^(1/3) .

Now I will use the Second derivative test:
f´(x) = 2.x - 1/x^2 = 2x - x^(-2)
then
f"(x) = 2 + 2.x^(-3) = 2 + 2/x^3

Now I will evaluate this second derivative at the point x0 =
(1/2)^(1/3) ;
f"(x0) = 2 + 2/[(1/2)^(1/3)]^3 = 2 + 2/(1/2) = 2 + 4 = 6 > 0 .
f"(x0) > 0 then f has a relative minimum at x0 = (1/2)^1/3 and this
point is the only one critical point.


I did it based in my own knowledge.
I hope this helps, but if you need any clarification, fell free to
post a request for an answer clarification.

Best regards.
livioflores-ga

Thanks You that helps alot !