00:01
In this problem we are provided with the function f of x which equals to 9 plus 12 times x squared minus 8 times x cubed.
00:13
So here we are asked to find out the local maxima as well as the local minima by making use of the first and the second derivative test.
00:26
So first we do so using the first derivative test.
00:32
So let us differentiate f with respect to x, we get 0 plus 12 times the derivative of x squared, which is 2 times x, minus 8 times the derivative of x cubed, which is 3 times x squared.
00:47
So simplifying this, we have 24 times x minus 24 times x squared.
00:54
So next we equate this first derivative to 0 and we solve for x to obtain the critical points.
01:02
So here we get 24 x times 1 minus x equals to 0 which clearly gives x equals to 0 and x equals to 1 to be the critical numbers.
01:19
So now using these two values we can split the interval as negative infinity up to 0, 0 up to 1 and 1 up to infinity.
01:32
So now considering any point in this interval and substituting it in the first derivative, we get f prime of negative 1 to be equal to 24 times negative 1 times 2 and clearly this is less than 0.
01:49
Next we have f prime of 0 .5 which equals to 24 times 0 .5 times 0 .5 which is clearly greater than 0 .5 .5.
02:02
Next we have f prime of 2 which equals to 24 times 2 times negative 1 and this is clearly less than 0.
02:13
So here we can see that f prime changes from less than 0 to greater than 0 which implies that there exists a local minimum at x equals to 0 and since here we see that it changes from greater than 0 to less than 0 to less than 0.
02:33
0, it implies that there exists a local maximum at x equals to 1.
02:40
So therefore, these are the required answers that are obtained using the first derivative test.
02:48
Next, let us make use of the second derivative test.
02:52
So let us differentiate f prime of x with respect to x again.
02:57
We get f double prime of x to be equal to 24 times 1 minus 24 times 2x...