Give an example of a rational function (i. e., the quotient of two polynomials) f satisfying the following conditions: • f is not defined at 1. • f(−3) = 0. • f(3) = 9. • lim x→5+ f(x) = −[infinity] and lim x→5− f(x) = [infinity]. explain your reasoning.

6x+18/(-x²+6x-5)

Step-by-step explanation:

• f is not defined at 1

• f(−3) = 0

• f(3) = 9

• lim x→5+ f(x) = −[infinity]

• lim x→5− f(x) = [infinity]

• f is not defined at 1

we need to have a denominator 0 for x = 1

so, 1/(x-1)

• lim x→5+ f(x) = −[infinity]

• lim x→5− f(x) = [infinity]

For the limits, we need to have

1/(-x+5)

this way, lim  x→5+ f(x) = −[infinity] and lim x→5− f(x) = [infinity]

So far we have

1/(x-1)(-x+5)

• f(−3) = 0

the nominator has to be 0 when x = -3

this way, x+3

so, (x+3)/(x-1)(-x+5)

• f(3) = 9

All we need to do is multiply (x+3)/(x-1)(-x+5) by 6, so

6(x+3)/(x-1)(-x+5) = 6x+18/(-x²+6x-5)

answer: $34.31

step-by-step explanation:

$11.99 x 3 = $35.97

$35.97 - 10% = $32.37

$32.37 + 6% = $34.31

1. 3.4/4

2. 0.8/7

3. 0.5 x 3

4. 0.9 x 6

5. 10.8

6. 2.35

7. 0.567

8. 9.35

9. 0.02

10. 0.06

11. 0.16

12. 0.42

13. 0.63

14. 5.6

15. 8.1

16. 9.9

17. 0.28

18. 0.48

19. 0.27

20. 0.49

21. 0.45

22. 0.72

23. 0.9

24. 0.88

25. 0.75

26. 0.16

27. 0.328

28. 0.495

29. 0.72

30. 1.068