Find the points on inflection and discuss the concavity of the graph of the function?

I need to find the points on inflection and discuss the concavity of the graph of the function:





f(x)= (x+1)/(sqrt(x))





I know that points of inflection require setting the second derivative equal to zero and finding where it is undefined.





Here is my work so far...


f(x)=(x^(-1/2))(x+1)


f(x)=x^(1/2)+x^(-1/2)


f'(x)=1/2x^(-1/2) -1/2x^(-3/2)


f''(x)= -1/4x^(-3/2) +3/4x^(-5/2)





My problem is that when I find the zeros of f''(x) I get x=3. That makes sense. But when I see where it is undefined, it is undefined from (-infinity,0).





How can I use this information to find points of inflection and where the graph is concave up/down?





Thanks so much!!!

Find the points on inflection and discuss the concavity of the graph of the function?
f''(x)=-1/4x^(-5/2)(x+3)





So the set where f'' is undefined is (-∞,0]. f'' is also undefined in 0.





Now inflection points could be found in points where second derivative f'' is undefined BUT in which f is defined.


Since f is also undefined on (-∞,0], then you don't need to


concern about these would-be points for this problem.





It remains x=3, correct. f'' also changes signs when x passes 3 so 3 is point of inflection





Let us write again


f''(x)=-1/4x^(-5/2)(x+3)





You study intervals for signs


When x%26lt;3. f'' is negative so there is concave down, for x%26gt;3


f'' is positive so there is concave up


The best way to do that is to make a table with all the intervals(for first derivative also) where f,f',f'' change signs





lobosito-- How did you get the answer you got for the second derivative? What did I do wrong there?


You didn't do nothing wrong, I just continued what you got.


f''(x)= -1/4x^(-3/2) +3/4x^(-5/2)


Now take common factor(you take he smallest power)


1/4x^(-5/2)( x^1+3), since -3/2=-5/2+1