We have been given a function
. We are asked to find the interval on which function is increasing and decreasing.
(a). First of all, we will find the critical points of function by equating derivative with 0.








So
are critical points and these will divide our function in 3 intervals
.
Now we will find derivative over each interval as:


Since
, therefore, function is increasing on interval
.


Since
, therefore, function is decreasing on interval
.
Let us check for the derivative at
.


Since
, therefore, function is increasing on interval
.
(b) Since
are critical points, so these will be either a maximum or minimum.
Let us find values of f(x) on these two points.




Therefore,
is a local maximum and
is a local minimum.
(c) To find inflection points, we need to check where 2nd derivative is equal to 0.
Let us find 2nd derivative.






Therefore,
is an inflection point of given function.