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## tmagloire1 one year ago Given f '(x) = (x + 1)(6 + 3x), find the x-coordinate for the relative minimum on the graph of f(x). 0 −1 −2 None of these ---- http://prntscr.com/8bljw7

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1. anonymous

do you know what $$y=(x+1)(6+3x)$$ looks like?

2. tmagloire1

a parabola

3. tmagloire1

upward

4. tmagloire1

@satellite73

5. anonymous

a parabola that opens up, right and the zeros are?

6. tmagloire1

-1 and -2

7. anonymous

ok good so lets draw a quick picture

8. anonymous

|dw:1441157312300:dw|

9. anonymous

at $$x=-2$$ the derivative goes from being positive (above the x axis) to negative (below the x axis) that tells you the function goes from increasing to decreasing at $$-2$$ i.e. up and then down therefore $$-2$$ is a relative max |dw:1441157412113:dw|

10. tmagloire1

exactly what i drew. i got 1.5ish as a minimum but it wasn't an option so i wasn't sure if they were trying to trick me by putting -1 and -2

11. anonymous

no it has a relative math right at $$x=-2$$ by a similar argument it has a relative min at $$x=-1$$

12. anonymous

at any rate the derivative has degree two, so the original function has degree 3 with positive leading coefficient|dw:1441157542119:dw|, so you have a good idea that it looks like this

13. tmagloire1

you lost me..

14. anonymous

ok lets go slow then

15. anonymous

you know what the derivative looks like right?

16. tmagloire1

yes

17. anonymous

and you know (or are supposed to be learning) that if the derivative is positive, then the function is increasing, and if the derivative is negative, tehn the function is decreasing

18. tmagloire1

yes i learned that

19. anonymous

ok good now "positive" is a synonym for "above the x axis"

20. anonymous

|dw:1441157806983:dw|

21. anonymous

that means on the interval $(-\infty,-2)$ and $(-1,\infty)$the function is increasing

22. anonymous

not the derivative, the original function $$f$$ also on $$(-2,-1)$$ the function is decreasing

23. anonymous

it goes up until it gets to -2, then it goes down that means -2 gives a relative MAX it goes down until -1, then it goes back up again that means -1 gives a relative MIN

24. tmagloire1

Ohh okay so you just us eincreasing/decreasing rules/ i understand thanks

25. anonymous

yw

26. tmagloire1

Can you help me with my second question as well?

27. anonymous

sure but i didn't see it

28. anonymous

oh now i do take the derivative what do you get?

29. anonymous

i hope you get $t^2-15t+50$

30. tmagloire1

For the derivative I got :t^2-50t+50

31. tmagloire1

oops i meant 15

32. anonymous

ok so your job is to solve $t^2-15t+50<0$

33. anonymous

because if the derivative is negative, it is moving to the left

34. anonymous

fortunately for you this factors, so it should be easy

35. tmagloire1

t<5

36. anonymous

is it clear how to solve that, or no?

37. tmagloire1

t>10

38. anonymous

i think you solved for where it is positive

39. tmagloire1

How am I supposed to solve this then? The two intervals i got were those

40. anonymous

it is zero at 5 and 10 right?

41. anonymous

|dw:1441158937764:dw|

42. tmagloire1

yes i got this

43. anonymous

since this parabola opens up, it is negative between the zeros and positive outside of them i.e negative on $$(5,10)$$

44. tmagloire1

is moving left another name for negative?

45. anonymous

moving to the left = derivative is negative

46. anonymous

i.e. the velocity is negative

47. tmagloire1

so 5<t<10 would be correct?

48. anonymous

yes

49. tmagloire1

Thank you for explaining them to me! I appreciate it!

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