what are points of inflection for 4x^4-x^3+2. if anyone knows ,plx help me

Question

hartnn · Answer

To get the points of inflection for f(x), equate the double derivative to 0
$f''(x)=0$
can you find f"(x) ?

anonymous · Answer

i have got two values of x i.e 0 and 1. i think 1 is not inflection point . is it so?

anonymous · Answer

The second derivative changes sign at an inflection point, so 
$$f''(x) = 0$$
is not enough to say that x is an inflection point, but it is a start. This is because f''(x) could go to zero but not change sign.

hartnn · Answer

^yes,
and how did you get x=1 ??
x=0 is correct.

anonymous · Answer

by putting 2nd derivativ of x =0

hartnn · Answer

what did you get f'"(x)as

hartnn · Answer

i mean, f"(x) as ?

anonymous · Answer

i hav not not undrstood wt u r asking

hartnn · Answer

what did you get 2nd derivative of  4x^4-x^3+2 as ?

anonymous · Answer

to get values of x

hartnn · Answer

did you get $f''(x)=48x^2-6x=0$ ?

anonymous · Answer

yes

anonymous · Answer

i hav found my mistake .values of x r 0 n 1/8

campbell_st · Answer

just a quick point of order... 

find the 1st derivative... and then let it equal zero and solve for x

this gives the stationary points.... which may be maximums, minimums... or horizontal points of inflection... 

then you test all the stationary points in the 2nd derivative. 

with the 2nd derivative... let it equal zero and solve for x. 

if there is a solution, check either side of the solution for a change in concavity. 

this way you'll be sure to get all the points of inflection...

anonymous · Answer

lots ov thnx