00:01
Here we have to determine the interval where the curve y is concave downward.
00:15
And so for that, we are going to use the concept of concavity.
00:20
According to concavity, we know that y is concave downward whenever the second derivative is less than zero.
00:28
So basically we have to determine the second derivative of y and then determine the interval where the second derivative is less than zero.
00:36
So let's see how to determine the second derivative two of y.
00:39
And from that we can determine the interval where the function y is concave downward.
00:48
So first let's find the first derivative of y.
00:52
Observe that we are given the y equals the definite integral from 0 to x t squared divided by t squared plus t plus 2 t.
01:01
And so i'm going to write this y and this equals the definite integral from 0 to x.
01:08
I consider the function inside the definite integral as f of t and so this becomes f of t d t.
01:17
That is we are considering f of t is basically equals the function inside the definite integral.
01:25
That is t squared divided by t squared plus t plus two.
01:31
Now this is in the form of fundamental theorem of calculus part one.
01:36
So according to that we can find the.
01:39
Derivative of y according to fundamental theorem of calculus part one we write y prime which is the derivative of y and this basically equals f of x and so this equals since we know f of t equals this expression you can find f of x by replacing all t in f of t by x and so therefore this is going to be x squared divided by x squared plus x plus 2.
02:11
And so we have determined the derivative of y.
02:15
That is the first derivative of y is this expression x squared divided by x squared plus x plus 2.
02:23
But we need the second derivative so that we can determine the concave downward interval.
02:30
So let's use this first derivative 2.
02:34
Here we have y prime and this equals x squared divided by x squared plus x plus 2.
02:42
Now we find the second derivative by differentiating both sides with respect to x.
02:48
And so on the left side, we are going to get y double prime, which is the second derivative.
02:54
To find the derivative of this expression, this is basically a quotient expression.
03:00
We use the quotient rule of derivatives.
03:02
So i'm going to first put the denominator terms that is x squared plus x plus 2 and then multiply this with the derivative of the numerator term which is x squared.
03:14
So its derivative is 2x when we use the power rule of derivative and then we put minus according to the quotient rule of derivative.
03:22
Now we put the numerator term that is x squared and then multiply this with the derivative of the denominator expression.
03:30
So the derivative of x squared plus x plus 2 and that equals 2x plus 1.
03:38
And this should be all over the square of the denominator expression.
03:48
It is x squared plus x plus 2 quantity square.
03:55
Now let's simplify this.
03:57
So here we have y double prime which represents a second derivative 2.
04:02
Let's distribute this 2x into the all terms inside the.
04:06
First bracket.
04:07
So 2x times x squared is 2x cube plus 2x times x is 2x squared plus 2x times 2 is 4x and here i'm going to distribute this negative x squared.
04:22
So negative x squared times 2x is negative 2x cube.
04:26
Then negative x squared times 1 is negative x squared and this has to be all over x squared plus x plus 2 .2 .2 let's simplify this.
04:46
You see that we have 2x cube and then negative 2x cube.
04:50
These two gets cancelled and we have the like terms 2x squared, negative x squared.
04:56
And so that gives x squared and then we have the term 4x.
05:03
So this basically is x squared divided by x squared plus 4x divided by the denominator expression.
05:11
That is x squared plus 4 x plus 2 quantity squared.
05:21
Now i'm going to write the numerator expression in factor form.
05:26
So we are going to have the y double prime.
05:29
By factoring x and we do that, we get x times of x plus 4.
05:34
This is divided by x squared plus x plus 2 quantity squared.
05:41
So we determine the second derivative 2 of y.
05:46
And that equals this expression.
05:51
Now let's create a sign diagram for the secondary derivative to so that we will understand where the curve y is concave downward.
06:00
And for that, first i'm going to replace y double prime by zero so that i can determine the point where the second derivative is zero.
06:09
So this we plug in y double prime equals zero and solve for x.
06:15
And so doing that we get zero equals x times...