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Illusion [34]
3 years ago
9

Consider the function f ( x ) = − 2 x 3 + 27 x 2 − 108 x + 9 . For this function there are three important open intervals: ( − [

infinity] , A ) , ( A , B ) , and ( B , [infinity] ) where A and B are the critical numbers. Find A and B
Mathematics
1 answer:
Darya [45]3 years ago
3 0

Answer:

<em>A=3 and B=6</em>

Step-by-step explanation:

<u>Increasing and Decreasing Intervals of Functions</u>

Given f(x) as a real function and f'(x) its first derivative.

If f'(a)>0 the function is increasing in x=a

If f'(a)<0 the function is decreasing in x=a

If f'(a)=0 the function has a critical point in x=a

As we can see, the critical points may define open intervals where the function has different behaviors.

We have

f ( x ) = - 2 x^3 + 27 x^2 - 108 x + 9

Computing the first derivative:

f' ( x ) = - 6 x^3 + 54 x - 108

We find the critical points equating f'(x) to zero

- 6 x^3 + 54 x - 108=0

Simplifying by -6

x^2 -9 x +18=0

We get the critical points

x=3,\ x=6

They define the following intervals

(-\infty,3),\ (3,6),\ (6,+\infty)

Thus A=3 and B=6

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Given: Triangle ABC : triangle ADB. AB =24, AD =16 <br> Find: AC
RoseWind [281]

Answer:

<h2>AC = 36.01</h2>

Step-by-step explanation:

Given ΔABC and ΔADB, since both triangles are right angled triangles then the following are true.

From ΔADB, AB² = AD²+BD²

Given AB = 24 and AD = 16

BD² = AB² - AD²

BD² = 24²-16²

BD² = 576-256

BD² = 320

BD = \sqrt{320}

BD = 17.9

from ΔABC, AC² = AB²+BC²

SInce AC = AD+DC and BC² = BD² + DC² (from ΔBDC )we will have;

(AD+DC)² = AB²+ (BD² + DC²)

Given AD = 16, AB = 24 and BD = 17.9, on substituting

(16+DC)² = 24²+17.9²+ DC²

256+32DC+DC² =  24²+17.9²+ DC²

256+32DC = 24²+17.9²

32DC = 24²+17.9² - 256

32DC = 640.41

DC = \frac{640.41}{32}

DC = 20.01

Remember that AC = AD+DC

AC = 16+20.01

AC = 36.01

6 0
3 years ago
8(x-2) = 64
Amiraneli [1.4K]

Answer:

I think d

Step-by-step explanation:

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5 0
2 years ago
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Add 9, 6 times
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3 years ago
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G = f - d / v what is f
Ratling [72]

Answer:

f = g + d/v

Step-by-step explanation:

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agasfer [191]

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