00:01
In this problem, we are asked to use differentials to approximate the value of 2 .99 the whole cube and then compare the result with that of the calculator.
00:13
Now, we can use differentials to approximate this.
00:18
For that, take f of x to be the function x cube.
00:24
Now, for a function f of x and a point x is equal to a, we can apply.
00:32
Approximate this function f of x at the point a using differentials as f of a plus the derivative of the function with respect to x at the point a times x minus a and this is known as the linear approximation of the function at the point so here take f of x to be the function x cube now we need to find f of 2 .9 or approximate f of 2 .99 then since f of 2 .99 will then become 2 .99 the whole cube which is the value that we require.
01:15
Now since 2 .99 is very close to 3 we can take the point about which we approximate to be x is equal to 3.
01:24
Then we have f of x will be approximately equal to f of 3 plus the derivative of f at the point 3 times x minus 3.
01:34
Using differentials.
01:37
Now find f of 3 which is obtained by substituting x is equal to 3 into the function.
01:43
It is 3 cube and 3 cube is 27.
01:47
Find f dash of x which is the first derivative of the function with respect to x and that is 3x square...