When is the function concave up/concave down. Already found Inflection points?

Question

anonymous · Answer

(A). f(x)=x^3/(x^2-1)
Inflection point x=0

(B). f(x)=x^4/(x^2-1)
No inflection point

campbell_st · Answer

ok... stationary points are found in the 1st derivative. 

the nature of the stationary point is x = a

f''(a) > 0  then a minimum occurs

f"(a) = 0 a horizontal pt of inflexion.

f"(a) < 0 a maximum occurs

anonymous · Answer

ok..

campbell_st · Answer

you also need to test either side of the possible  point of inflexion to find a change of sign.

anonymous · Answer

what i usually did was plug numbers on either side of the inflection point i.e -1 and 1 (from part a) into the second derivative. but then the answer is really confusing

campbell_st · Answer

for A its a horizontal pt of inflexion

B is correct.

campbell_st · Answer

well you'll find x = 0 is a solution to f'(x) = 0... which means its a stationary point... but you don't know what type (max, min or pt of inflexion)

so use the 2nd derivative test... 

and you'll find f"(0) = 0 so its a horizontal point of inflexion (special case)

checking either side of the solution to f"(x) = 0 can be used.

anonymous · Answer

i know how to find the inflection point. I just want to know where the function is concave up or down. for the (b) part there are no inflection points. so i don't know how to proceed.

anonymous · Answer

@campbell_st

campbell_st · Answer

there is a local max and min... but you need to find those values from the 1st derivative

I got

$$f'(x) = \frac{x^2(x^2 -3)}{(x^2 - 1)^2}$$

so stationary points occur at 

$$x = 0, \pm \sqrt3$$

then $$f"(x) = \frac{2x(x^2 + 3)}{(x^2 - 1)^3}$$

so x = 0 is a horizontal pt of inflexion since f"(0) = 0

$$f(\sqrt{3}) > 0 $$ so a min

$$f"(-\sqrt{3}) < 0 $$ so a max