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Angelina_Jolie [31]
3 years ago
14

Find all critical numbers for the following function. Then use the​ second-derivative test on each critical number to determine

whether it leads to a local maximum or minimum.

Mathematics
1 answer:
Gelneren [198K]3 years ago
5 0
Critical values are values where f'(x)=0 and the bounds of a function. Thus, let's solve for f'(x)!

f(x)=2x^3-3x^2+3x+8
f'(x)=6x^2-6x+3

Now let's set f'(x)=0

0=6x^2-6x+3
0=2x^2-2x+1

As it turns out, 2x^2-2x+1 isn't factorable!
This saves me some time because this means there are no critical numbers!
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Answer:

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Step-by-step explanation:

We are given that consider a game in which players roll a number cube to determine the number of points earned. If a player rolls a prime number, that many points will be added to the player’s total. Any other roll will be deducted from the player’s total.

Assuming that the numbered cube is a dice with numbers (1, 2, 3, 4, 5, and 6).

Here, the prime numbers are = 1, 2, 3 and 5

Numbers which are not prime = 4 and 6

This means that if the dice got the number 1, 2, 3 or 5, then that many points will be added to the player’s total and if the dice got the number 4 or 6, then that many points will get deducted from the player’s total.

Here, we have to make a probability distribution to find the expected value of the points earned on a single roll in this game.

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                      +1                                                 \frac{1}{6}

                      +2                                                \frac{1}{6}

                      +3                                                \frac{1}{6}

                      -4                                                 \frac{1}{6}

                      +5                                                \frac{1}{6}

                      -6                                                 \frac{1}{6}

Here (+) sign represent the addition in the player's total and (-) sign represents the deduction in the player's total.

Now, the expected value of X, E(X)  =  \sum X \times P(X)

   =  (+1) \times \frac{1}{6} +(+2) \times \frac{1}{6} +(+3) \times \frac{1}{6} +(-4) \times \frac{1}{6} +(+5) \times \frac{1}{6} +(-6) \times \frac{1}{6}

   =  \frac{1}{6} + \frac{2}{6} + \frac{3}{6} - \frac{4}{6} + \frac{5}{6} - \frac{6}{6}

   =  \frac{1+2+3-4+5-6}{6}

   =  \frac{11-10}{6}= \frac{1}{6}

Hence, the expected value of the points earned on a single roll in this game is  \frac{1}{6} = 0.1667 .

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