00:01
Hello, students, we are given a function.
00:01
We just need to find the intervals on which f is increasing or decreasing, and we just need to find the local maximum minimum values and the intervals of concavity and inflection point.
00:10
So what we are given, fx is equals to x divided by x squared plus 1.
00:16
Okay, now we will go for the first part, a.
00:18
We are asked for the increasing.
00:20
For increasing, we are supposed to know that f dash x is always greater than 0.
00:26
Okay, so first of all, we will have to find f dash x.
00:29
How we will calculate it so basically it will be x square plus one all the denominator and upper part will be x squared plus one times the derivative of x is nothing but one minus x times of it will be 2x okay after that how we will solve it f dash x is equal to here x square plus 1 minus 2x square divided by here x square plus 1 whole square so basically f dash x comes out as 1 minus x square minus 2x square comes out x xx is 1 okay and after that x square plus one whole square now we are supposed to know that f dash x have to equal to sorry greater than zero okay so one minus x squared divided by x square plus one whole square is equal to sorry greater than zero and the denominator part will always be positive because it is in the whole square form so one minus x square can be written as zero greater than zero so x square minus one less than zero it means x belongs to minus one to one to one okay where is in the increasing form okay okay now, decreasing, for decreasing, we are supposed to know that f -dash -x have to be less than 0.
01:36
So basically, we can write that again, we can directly solve that 1 minus x square is less than 0.
01:43
It means x square minus 1 is greater than 0.
01:46
And x plus 1 times x minus 1 is greater than 0.
01:51
It means x belongs to minus infinity to minus 1, union 1 to infinite.
01:56
Okay now we will go for the b part what we are asked that the local max and minimum values okay so f dash x was nothing but f dash x is nothing but equal to one minus x squared divided by x square plus 1 whole square okay you can check again 1 minus x squared divided by x square plus 1 all square in order to find those values f to be equal to 0 it means we can directly say that 1 minus x square is equal to 0 it means x square comes out as one x comes out as plus minus one now what we need to do we just need to check for f double dash x okay so f double dash x sorry it will be we can directly write it f double dash x okay so basically what we can here solve x square plus one whole square and the whole square okay so basically we can write x square plus one whole square times off see differentiation of 1 minus x squared is nothing but minus 2x okay minus of here 1 minus x square times of the derivation of 1 x squared plus 1 always score is nothing but 2 times of 1 plus x square okay which is nothing but x squared plus 1 and times of 2x okay students so basically what we can conclude that f double dash x comes out as minus of here 2x times x square plus 1 whole square okay and after that minus of here 2 2 j 4 okay so basically we can just write it as minus of 2 2 j4 and x times of 1 minus x squared times 1 plus x square is nothing but 1 minus x to the power 4 okay and divided by here x square plus 1 whole power 2 2 2 j 4 okay now we will check those values for x equal to 1 so basically minus 2 times 1 square is 1 1 2 2 square is nothing but 4 minus 4 times 1 4 okay after that 1 minus 1 0 0 0 0 0 after that, a1 plus 1 is equal to 2, 2 to the power 4 comes out as 16.
03:54
So basically it will be 0.
03:56
It will be minus of 8 divided by 16 comes out as minus 1 by 2, which is less than 0.
04:00
It means at x equal to 1, x equal to 1, f will be maximum.
04:06
Okay? f has a maximum value.
04:08
Okay, so we can just find the maximum value.
04:12
By putting x equal to 1 in the function, what was our function? x divided by x squared plus 1.
04:17
It means we can write x, it means 1 square is 1 plus 1.
04:21
1 by 2 okay okay it is nothing but the y max now so basically we can write local maximum point local maximum is nothing but 1 comma 1 by 2 okay students now what we need to do we will go with the x equal to minus 1 so basically when we put the f double dash of minus 1 we will get here what we will get minus 2 times of minus 1 comes out as 2 okay and 2 times of here it is 1 plus 1 2 so 1 square 2 square is nothing but 4 so 2 4 2 times of 4 minus 4 times of okay students and after that here 1 minus it will be 1 minus 1 okay it will be 0 divided by here 1 minus 1 all square is 1 plus 1 and whole power 4 so basically 8 it will be 0 and divided by 16 comes out as 1 by 2 which is greater than 0 it means x equal to minus 1 y will be minimum okay local minimum so basically we can just put that value f of minus 1 comes out as minus 1 divided by minus 1 all the score is 1 plus 1 is equal to minus 1 by 2 so basically local what we can write local minimum okay which is point is nothing but minus 1 comma minus 1 by 2 now we will go for the c part find the intervals on the concavity and the inflection points okay in order to find the intervals of concavity we just need to find the f double dash x okay so basically first of all we will write here again f dash x what was our f dash x we can check again it was one minus sorry one minus x squared divided by x square plus one one minus x square divided by x square whole square okay now we have already calculated the f double dash x value also so basically we can check again a double dash x it was minus 2 x x squared plus 1 all square okay so basically we can write minus 2 x times x square plus 1 whole square minus 4x times 1 minus x to the power 4 okay we can check again minus 2x times x square plus 1 or square minus 4x times 1 minus x to the power 4 divided by divided by x square plus 1 whole power 4 okay for concave concave we just need to know that f double dash x have to be greater than zero.
06:54
So basically when we put that, it will be automatically zero...