00:01
In this problem we are provided with the function f of x equals to natural log of square root of x squared plus 1 and we need to find out the interval where this function is decreasing.
00:21
So for that we make use of the first derivative test.
00:25
Let us differentiate the given function.
00:27
We have the derivative of natural log of square root of x squared plus 1 to be 1 over square root of x squared plus 1 times differentiating square root of x squared plus 1 we get 1 over 2 2 2 2 times 1 1 over square root of x squared plus 1 which is 2x so here 2 and 2 cancel out and we can write f prime of x to be equal to x over square root of x squared plus 1 times square root of x squared plus 1 is x squared plus 1.
01:05
So next we equate this first derivative to 0 and on doing so we obtain x to be equal to 0.
01:16
So now we can split the number line into two intervals that is from negative infinity up to 0 and from 0 up to positive infinity.
01:30
Now let us check whether the function is increasing or decreasing in these intervals.
01:35
So first let us consider the interval negative infinity up to 0.
01:39
So here we can evaluate the value of f prime at a negative 1 and on substituting the value of x as negative 1 in f prime of x, we get negative 1 over 2 which is less than 0.
01:52
So this implies that the function f is decreasing in this interval...