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Alborosie
3 years ago
7

Find the absolute extrema if they​ exist, as well as all values of x where they​ occur, for the function f (x )equals2 x Supersc

ript 4 Baseline minus 36 x squared plus 2 on the domain [negative 4 comma 4 ].
Find the derivative of f left parenthesis x right parenthesisequals2 x cubed plus 20 x squared plus 50 x plus 1. f prime left parenthesis x right parenthesisequals nothing Identify the absolute maximum if it​ exists, as well as all values of x where it occurs.
Mathematics
1 answer:
kkurt [141]3 years ago
5 0

Step-by-step explanation:

Derivative of the first function:

f(x)=2x^{4} -36x^{2} +2  on D:[-4,4]

f'(x)=8x^{3} -72x

set that equal to 0 and solve for possible x values.

8x(x^{2} -9)=0\\x=0, -3, 3

put x back in to original equation to get a y value

y = 2, -160, -160,

Absolute min at (-3, -160) and (3, 160) and an absolute max at (0,2)

Second Part

Find derivative of f(x)=2x^{3} +20x^{2} +50x +1

f'(x)=6x^{2} +40x+50

set equal to zero, it is a quadratic so you can use the quadratic formula to solve for x

x= -5 or -5/3

put x back into the original function to get a y value

y = 1 or -973/27

so absolute max at (-5, 1) and absolute min at (-5/3, -973/27)

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In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini
svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2

The augmented matrix is

\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]

The reduction of this matrix to row-echelon form is outlined below.

R_2\rightarrow R_2-R_1

\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]

R_3\rightarrow R_3-2R_1

\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]

R_3\rightarrow R_3-R_2

\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]

The last row determines, if there are solutions or not. To be consistent, we must have k such that

k^2-k-2=0

\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2

Case k = −1:

\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]

This gives the infinite many solution.

5 0
3 years ago
I NEED HELP (I’m sooo dumb lol)
statuscvo [17]
To make a box and whisker plot, first you write down all of the numbers from least to greatest.

0, 1, 3, 4, 7, 8, 10

The median is 4, so that’s the middle line of the plot.

So now we have:
0, 1, 3, [4,] 7, 8, 10

So next we have to find the 1st and 3rd interquartiles..
0, [1,] 3, [4,] 7, [8,] 10

Those are the next 2 points you put on the plot.

Lastly, the upper and lower extremes. These are the highest and lowest numbers in the data.
[0,] 1, 3, 4, 7, 8, [10]

These are the final points on the plot.

To make the box of a box-and-whisker plot, you plot the 3 Medians of the data: 1, 4, and 8, and connect those to make a box that has a line in the middle at 4.

Next, you plot the upper and lower extremes: 0 and 10, by making “whiskers” that connect to the box. So you draw a line from the extremes to the box.
7 0
3 years ago
A package of nine 12-ounce cans of cola costs $7.50. A half-gallon bottle of the same cola is sold for $2.90. Which cola has a c
goldfiish [28.3K]

Answer:

Cola bottles are cheaper. Mark me brainliest!

Step-by-step explanation:

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2 years ago
Evaluate 29 • 1 + 2(20 ÷ 4 – 5)
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Writing this to meet 20 character requirement
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2 years ago
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