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myrzilka [38]
3 years ago
12

Hey can you please help me posted picture of question

Mathematics
2 answers:
svet-max [94.6K]3 years ago
4 0
If we observe the graph we can see that the curve crosses the x-axis at the following points: 1, 3, and 5
This means x = 1, x = 3 and x = 5 are the roots of the polynomial and x - 1, x - 3 and x - 5 are the factors of the polynomial.

We can express the polynomial as the product of its factors. So the polynomial will be (x-1)(x-3)(x-5)

The correct answer to this question is option A

Charra [1.4K]3 years ago
4 0
The correct Option is (A) (x-1)(x-3)(x-5)

Explanation:
As you can see in the graph that the graph intersects the x-axis at 3 points which are:

x = 1;
x = 3;
x = 5.

So the correct polynomial product is:
(x-1)(x-3)(x-5)

Because if we put them equals to zero we would get:
x-1=0
=> x=1

x-3 = 0
=> x = 3

x-5 = 0
=> x = 5(same as stated above)

Hence the correct option is (A) (x-1)(x-3)(x-5)
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Consider the function f ( x ) = − 2 x 3 + 27 x 2 − 108 x + 9 . For this function there are three important open intervals: ( − [
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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

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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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