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Ulleksa [173]
4 years ago
7

F(x)=-x^(4)-9x^(3)-24x^(2)-16x what is the relative maxium and minimum?

Mathematics
1 answer:
stepladder [879]4 years ago
7 0
We'll need to find the 1st and 2nd derivatives of F(x) to answer that question.

F '(x) = -4x^3 - 27x^2 - 48x - 16     You must set this = to 0 and solve for the 
                                                           roots (which we call "critical values).

F "(x) = -12x^2 - 54x - 48

Now suppose you've found the 3 critical values.  We use the 2nd derivative to determine which of these is associated with a max or min of the function F(x).

Just supposing that 4 were a critical value, we ask whether or not we have a max or min of F(x) there:   

F "(x) = -12x^2 - 54x - 48    becomes    F "(4) = -12(4)^2 - 54(4)
                                                                        =  -192 - 216
                                       Because F "(4) is negative, the graph of the given
                                        function opens down at x=4, and so we have a 
                                        relative max there.  (Remember that "4" is only
                                       an example, and that you must find all three
                                        critical values and then test each one in F "(x).
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Peter the optician sees 16 patients one Tuesday.4 are neither short nor long sighted,the total number of short sighted patients
My name is Ann [436]

The number of short-sighted patients is 7

The number of long-sighted patients is 5

The number of patients requiring a varifocal lens is 4

Step-by-step explanation:

Stepwise solution-

Throughout the question, only 3 types of patients are mentioned. Hence, it would be safe to assume that he saw/sees these 3 types of patients on Tuesday.

As per the given details-

4 are neither short-sighted patients nor long-sighted patients. Then it means no of varifocal lens patients is 4.

Hence remaining total patients=16-4=12

As given, no of short-sighted patients is 2 more than the number of long-sighted patients.

Let us assume the number of long-sighted patients as “x”

Thus, the number of short-sighted patients would be “x+2”

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5 0
4 years ago
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nika2105 [10]

Are you simplifying the equation or solving for x?

Finding x:

You first expand the three terms in parenthesis, you do this by taking

5(x - 1) and multiplying 5 by x and -1 which gives 5x - 5, you repeat this for 6(x + 2), -7(x - 3) and 8(x + 1) which gives 6x + 12, -7x + 21 and 8x + 8

Collectively we have the equation 5x - 5 + 6x + 12 - 7x + 21 = 8x + 8

Next we group the like terms

5x + 6x - 7x

5 + 6 = 11 and 11 - 7 = 4, thus we have 4x and -5 + 12 + 21 = 28

Now the equation has been simplified to 4x + 28 = 8x + 8

Then we subtract each side by 28 which is 4x = 8x - 20

Then 8x is subtracted from each side, giving -4x = -20

Lastly both sides are divided by -4 to isolate x which gives x = 5

Fully simplified:

To have the equation fully simplified you just stop at 4x + 28 = 8x + 8 if it is an answer choice, if it is not factor the equation.

I will be using GFC (Greatest Common Factor) to factor.

We start factoring at turning 28 into 4 * 7

4x + 28 = 4x + 4 * 7

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Repeat for 8x + 8, where 8x + 8 = 8 * 1, the common factor is 8 so you remove it giving the final simplification of:

4( x + 7 ) = 8( x + 1 )

5 0
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