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Brut [27]
3 years ago
9

Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = 6x3 − 9x2 − 108x + 6, [−3, 4]

Mathematics
2 answers:
Leni [432]3 years ago
8 0

Answer:

The absolute maximum of f(x) on [-3, 4] is 138 and the absolute minimum of f(x) on [-3, 4] is -237.

Step-by-step explanation:

To find the absolute extrema values of f(x) = 6x^3 - 9x^2 - 108x + 6  on the closed interval [−3, 4] you must:

1. Locate all critical values. We need to find the derivative of the function and set it equal to zero.

\frac{d}{dx}f(x)= \frac{d}{dx}\left(6x^3-9x^2-108x+6\right)=\\\\f'(x)=\frac{d}{dx}\left(6x^3\right)-\frac{d}{dx}\left(9x^2\right)-\frac{d}{dx}\left(108x\right)+\frac{d}{dx}\left(6\right)\\\\f'(x)=18x^2-18x-108

18x^2-18x-108=0\\18\left(x^2-x-6\right)=0\\18\left(x+2\right)\left(x-3\right)=0\\\\\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}\\x=-2,\:x=3

2. Evaluate f(x) at all the critical values and also at the two values -3 and 4

\left\begin{array}{cc}x&f(x)\\-3&87\\-2&138\\3&-237\\4&-186\end{array}\right

3. The absolute maximum of f(x) on [-3, 4] will be the largest number found in Step 2, while the absolute minimum of f(x) on [-3, 4] will be the smallest number found in Step 2.

Therefore,

The absolute maximum of f(x) on [-3, 4] is 138 and the absolute minimum of f(x) on [-3, 4] is -237.

Alborosie3 years ago
5 0
It is convenient to use a graphing calculator for this. The graph shows the maximum and minimum are not at the ends of the interval, so could be found by differentiating the function and setting that derivative to zero. The function would then need to be evaluated for those solutions {-2, 3}.

The absolute minimum on the interval is -237 at x=3.
The absolute maximum on the interval is 138 at x=-2.

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Anyone know how to solve this step by step?
ira [324]

Answer:

-1/2 or x=2

Step-by-step explanation: The first step you need to do in getting your answer is factoring ,you will factor the left side of the equation first:

(2x+1)(x-2)=0

Next, you will set the factors to equal to "0":

2x+1=0  x-2=0

Then you'll proceed to solve the problem using the basic algebraic equation problem solving steps .When you do this you'll be left with answers of x=-1/2  and x=2


3 0
3 years ago
-
bixtya [17]

Using an exponential function, it is found that f(5.5) = 19.8.

<h3>What is an exponential function?</h3>

An exponential function is a function in which the growth rate is a percentage, modeled by:

y = ab^x

In which:

  • a is the initial value.
  • b is the rate of change.

f(3.5) = 25 means that:

ab^{3.5} = 25

a = \frac{25}{b^{3.5}}

f(8.5) = 14 means that:

ab^{8.5} = 14

Hence:

\frac{25}{b^{3.5}} \times b^{8.5} = 14

25b^5 = 14

b = \sqrt[5]{\frac{14}{25}}

b = 0.89

a = \frac{25}{0.89^{3.5}} = 37.59

Hence, the function is given by:

y = 37.59(0.89)^x

Then, when x = 5.5:

f(5.5) = 37.59(0.89)^{5.5} = 19.8

More can be learned about exponential functions at brainly.com/question/25537936

7 0
2 years ago
Assume that a policyholder is four times more likely to file exactly two claims as to file exactly three claims. Assume also tha
kykrilka [37]

Answer:

1.3125

Step-by-step explanation:

Given that our random variable X follows a Poisson distributionP(X=k)=\frac{\lambda^k e^-^p}{k!} \ \ \ \ \ \ \ \ p=\lambda

Evaluate the formula at k=2,3:

P(X=2)=0.5\lambda^2e^{-\lambda}\\\\P(X=3)=\frac{1}{6}\lambda^2e^{-\lambda}

#since

4P(X=2)=P(X=3);\\\\0.5\lambda^2e^{-\lambda}=4\times\frac{1}{6}\lambda^3e^{-\lambda}\\\\0.5\lambda^2=\frac{2}{3}\lambda^3\\\\0.75=\lambda

The mean and variance of the Poisson distributed random variable is equal to \lambda:

\mu=\lambda=0.75\\\sigma ^2=\lambda=0.75

#By property variance:

\sigma ^2=V(X)=E(X^2)-(E(X))^2=E(X^2)\\\\E(X^2)=\sigma^2+\mu^2=0.75+0.75^2=1.3125

The expectation is 1.3125

4 0
3 years ago
What is the length of AC?
natima [27]

7/x = 84 / (156 -x)
84x = 7(156 -x)
84x = 1092 - 7x
91x = 1092
x = 12

AC = 156 - 12 = 144

answer
D. 144
3 0
3 years ago
Please please help me!!
ElenaW [278]
Hey you can go on Gauthmath it will have more people online to help you
8 0
3 years ago
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