At the start of his work day, Mr. Eriksen had 20 boxes of light bulbs tested and ready to be shipped. On average, he can test 40
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I assume otherwise. If
is indeed a proper PDF, then its integral over the support of
is 1.
Compute the integral.
Then
First of all, when I do all the math on this, I get the coordinates for the max point to be (1/3, 14/27). But anyway, we need to find the derivative to see where those values fall in a table of intervals where the function is increasing or decreasing. The first derivative of the function is 
. Set the derivative equal to 0 and factor to find the critical numbers. 
, so x = -3 and x = 1/3. We set up a table of intervals using those critical numbers, test a value within each interval, and the resulting sign, positive or negative, tells us where the function is increasing or decreasing. From there we will look at our points to determine which fall into the "decreasing" category. Our intervals will be -∞<x<-3, -3<x<1/3, 1/3<x<∞. In the first interval test -4. f'(-4)=-13; therefore, the function is decreasing on this interval. In the second interval test 0. f'(0)=3; therefore, the function is increasing on this interval. In the third interval test 1. f'(1)=-8; therefore, the function is decreasing on this interval. In order to determine where our points in question fall, look to the x value. The ones that fall into the "decreasing" category are (2, -18), (1, -2), and (-4, -12). The point (-3, -18) is already a min value.