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Katena32 [7]
2 years ago
8

Please help me with this problem. NO LINKS!!!

Mathematics
1 answer:
Westkost [7]2 years ago
5 0

Answer:

  c. x about 57/16

Step-by-step explanation:

You have not specified the algorithm you use for one iteration. We will define it as follows:

The starting point is a pair of numbers that are upper and lower bounds on the solution. The iteration ends when a new value replaces one of these bounds. The final estimate of the root will be the average of the latest upper and lower bounds.

__

The given graph shows suitable initial bounds are 3 and 4. The sign of h(x) = f(x)-g(x) matches the sign of h(3) when x=7/2, so the value 7/2 will replace the lower bound at the end of iteration 1.

The new average of upper and lower bounds is (7/2 +4)/2 = 15/4. The sign of h(15/4) matches the sign of h(4), so 15/4 becomes the new upper bound at the end of iteration 2.

The new average of the bounds is (7/2 +15/4)/2 = 29/8. The sign of h(29/8) matches the sign of h(15/4), so 29/8 becomes the new upper bound at the end of iteration 3.

After 3 iterations, the bounds are 7/2 and 29/8. The average of these values is the approximate solution to the equation:

  x = (7/2 +29/8)/2 = 57/16

_____

<em>Additional comment</em>

We have tried to be clear about what we consider to be one iteration, and how a root approximation is arrived at. The definition of these things provided by your curriculum materials may be different.

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konstantin123 [22]

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3 years ago
Select ALL the correct answers.<br> Select all functions that have a y-intercept of (0,5).
otez555 [7]
1.) f(x)=7(b)^x-2
x=0→f(0)=7(b)^0-2=7(1)-2=7-2→f(0)=5→(x,f(x))=(0,5) Ok

2.) f(x)=-3(b)^x-5
x=0→f(0)=-3(b)^0-5=-3(1)-5=-3-5→f(0)=-8→(x,f(x))=(0,-8) No

3.) f(x)=5(b)^x-1
x=0→f(0)=5(b)^0-1=5(1)-1=5-1→f(0)=4→(x,f(x))=(0,4) No

4.) f(x)=-5(b)^x+10
x=0→f(0)=-5(b)^0+10=-5(1)+10=-5+10→f(0)=5→(x,f(x))=(0,5) Ok

5.) f(x)=2(b)^x+5
x=0→f(0)=2(b)^0+5=2(1)+5=2+5→f(0)=7→(x,f(x))=(0,7) No

Answers:
First option: f(x)=7(b)^x-2
Fourth option: f(x)=-5(b)^x+10 
7 0
3 years ago
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