00:01
Hello everyone, in this question we have to find the value of x for which the tangent line is horizontal.
00:06
For that find f ' of x.
00:08
F ' of x is 6x5 x -13 7 plus x to the power 6 7 into x -13 6.
00:19
So taking common x to the power 5 and x -13 to the power 6 we will be having 6 into x -13 plus 7x.
00:33
So calculating this x -13 to the power 6 6x -78 plus 7x.
00:43
So this gives you x to the power 5 x -13 to the power 6 13x -78.
00:51
So taking 13 common we will be having x -6.
00:59
So therefore 13x to the power 5 x -13 to the power 6 x -6.
01:06
So equating f ' of x equal to 0 we will be getting x as 0, x as 13 and x as 6.
01:15
So for this points the tangent value is horizontal.
01:19
Hello everyone, in this question f of x is given.
01:24
Similarly f ' of x is 1 by 2 square root of x square minus 14x plus 63 into 2x minus 14.
01:35
So this will become 2 into x minus 7 divided by 2 into square root of x square minus 14x plus 63.
01:43
So this 2 and this 2 will get cancelled.
01:46
Equating f ' of x equal to 0 we will be getting x equal to 7.
01:50
So at x equal to 7 the tangent line is horizontal.
01:55
Next the third part is the domain of g is equal to question mark.
02:03
That is they have given that g of x is equal to square root of 6x.
02:09
So from this we can say that x can only be positive real numbers.
02:22
That is domain of g is greater or equal to 0.
02:36
The next part is they have given f of u equal to log u.
02:42
So we have to find the domain of f.
02:47
Here domain of f cannot be negative numbers.
03:01
So domain of f is equal to positive real numbers.
03:13
So domain of f is greater than or equal to 0.
03:20
Similarly they have given g of x is equal to 49 minus x square.
03:28
Similarly the domain of g can all be real numbers.
03:44
So domain of g equal to real numbers...