
now, we get critical points from zeroing out the derivative, and also from zeroing out the denominator, but those at the denominator are critical points where the function is not differentiable, namely a sharp spike or cusp or an asymptote.
so, from zeroing out the derivative we get no critical points there, from the denominator we get x = 8, but can't use it because f(x) is undefined.
therefore, we settle for the endpoints, 4 and 6,
f(4) =3 and f(6) = 7
doing a first-derivative test, we see the slope just goes up at both points and in between, but the highest is f(6), so the absolute maximum is there, while we can take say f(4) as the only minimum and therefore the absolute minumum as well.
| Red |Green | Brown Total
<span>Smile |0.100 | .300 | .10 0.50</span>
<span>No smile |0.200 | .150 | .15 0.50
Total 0.30 0.45 0.25 1.00
The probability of choosing red candy is 0.10
The probability of choosing no smile candy is 0.50
The probability of choosing red or no smile = 0.10 + 0.50 = 0.60
I did not use 0.30 as the probability of red because it included the no smile candy. It should only be red candy that has a smile.</span>
SA=4pir^2
d/2=r
8/2=4
r=4
SA=4pi4^2
SA=4pi16
SA=64pi in^2
if want, aprox pi=3.14
SA=200.96 in^2