## anonymous one year ago Precalculus help!! 1. Find the derivative of f(x) = 6 divided by x at x = -2. and 2. Find the derivative of f(x) = 4x + 7 at x = 5. Thank you. Need help solving, will medal.

1. freckles

I bet you are practicing using the formal definition and not the short cuts right?

2. anonymous

Yes @freckles

3. freckles

you could use either one of these: $f'(a)=\lim_{h \rightarrow 0}\frac{f(a+h)-f(a)}{h} \text{ or } f'(a)=\lim_{x \rightarrow a} \frac{f(x)-f(a)}{x-a}$

4. freckles

they are the same definition just written a little different

5. freckles

which one are you more comfortable with?

6. anonymous

I'm not sure, I'm still confused with both.

7. freckles

Ok we could do question one together both ways and you can decide which way you want to use for the second and I will check your work for that one

8. anonymous

Ok thanks!

9. freckles

$a=-2 \\ f(x)=\frac{6}{x} \\ f(a)=f(-2)=\frac{6}{-2}=-3 \\ f(a+h)=f(-2+h)=\frac{6}{-2+h}$ first do you understand how I got f(a+h) and f(a) for the first one?

10. anonymous

Yes completely.

11. freckles

Ok I'm going to input them into the formula

12. freckles

$f'(a)=f'(-2)=\lim_{h \rightarrow 0}\frac{\frac{6}{-2+h}-(-3)}{h}$

13. freckles

$f'(-2)=\lim_{h \rightarrow 0}\frac{\frac{6}{-2+h}+3}{h}$

14. freckles

ok the trick is to get that h on the bottom to cancel somehow so we can actually plug in h=0

15. freckles

notice we have a compound fraction

16. freckles

that is a fraction inside a bigger fraction

17. anonymous

Hold on super quick, writing this down.

18. anonymous

Ok got it.

19. freckles

we want to see if we can "simplify" the fraction that means get rid of the compound fraction notice the little fraction has denominator (-2+h) multiply top and bottom by (-2+h) this will make our compound fraction into a "simple" fraction do you want to try multiplying top and bottom by (-2+h)

20. freckles

and let me know what the result is just doing that one step

21. anonymous

Yes

22. anonymous

Do I multiply the top and bottom of just the little fraction?

23. freckles

multiply top and bottom by (-2+h) of the big fraction

24. anonymous

Would it be 9/h(-2+h) ?

25. freckles

that isn't a bad try ... I think the 9 is just a bit off...should be 3 there $\lim_{h \rightarrow 0} \frac{\frac{6}{-2+h}+3}{h} \cdot \frac{-2+h}{-2+h}=?$ $\lim_{h \rightarrow 0} \frac{\frac{6(-2+h)}{-2+h}+3(-2+h)}{h(-2+h)} \\ =\lim_{h \rightarrow 0}\frac{6+3(-2+h)}{h(-2+h)}$

26. freckles

3(-2+h)=-6+3h

27. freckles

$\lim_{h \rightarrow 0} \frac{6-6+3h}{h(-2+h)}$

28. anonymous

Oh ok, I see. So it would be 3h/h(-2+h)?

29. freckles

$\lim_{h \rightarrow 0}\frac{3h}{h(-2+h)}$ you should recall h/h=1 (we can say this since h is not 0)

30. freckles

$\lim_{h \rightarrow 0}\frac{3 \cancel{h}}{\cancel{h}(-2+h)}$

31. freckles

guess what ?

32. anonymous

So then we are left with 3/-2+h

33. freckles

we are going to have no problems pluggin in 0 for h now because this function 3/(-2+h) actually exists at h=0

34. freckles

so what do you think f'(-2)=?

35. anonymous

But how do you know that the function exists at 0?

36. freckles

because the bottom is not 0 when h is 0

37. anonymous

Oh ok got it, so would f'(-2) = 3/-2?

38. freckles

yes

39. freckles

or -3/2

40. freckles

doesn't matter where you put the negative at

41. anonymous

Could you check my other question once I solve it?

42. freckles

k.. while you are working it I'm going to do the first one again but use the other formula I wrote

43. anonymous

Ok (:

44. anonymous

Is the second one f(5) = 4?

45. freckles

$f'(a)=\lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a} \\ a=-2 \\ f(x)=\frac{6}{x} \\ f(a)=f(-2)=\frac{6}{-2}=-3 \\ f'(a)=f'(-2)=\lim_{x \rightarrow -2} \frac{\frac{6}{x}+3}{x+2} \\ \text{ multiply \top and bottom by } x \\ f'(-2)=\lim_{x \rightarrow -2} \frac{6+3(x)}{(x+2)(x)} \\ f'(-2)=\lim_{x \rightarrow a}\frac{3(x+2)}{(x+2)(x)} \\ \text{ notice} \frac{x+2}{x+2}=1 \text{ for when } x \neq -2 \\ f'(-2)=\lim_{x \rightarrow -2} \frac{3}{x} =\frac{3}{-2} \text{ or } \frac{-3}{2} \text{ or } -\frac{3}{2} \text{ or } -1.5$

46. freckles

oops one sec let me check

47. freckles

"2. Find the derivative of f(x) = 4x + 7 at x = 5." f(5)=4(5)+7=20+7=27

48. anonymous

Oh ok, I'm used to the first version we used though. But the one you just used also makes a lot of sense. (:

49. anonymous

Yes, and for f(a+h) I got it = to 27+h

50. freckles

did you mean f'(5)=4?

51. anonymous

Yes that's what I meant.

52. freckles

that is right

53. anonymous

Could you check this one too? The position of an object at time t is given by s(t) = -8 - 9t. Find the instantaneous velocity at t = 1 by finding the derivative.

54. freckles

yeah this question is basically the same as the last questions you asked just asked it in word problem form

55. anonymous

Would it be: f'(1) = 9?

56. freckles

the velocity=derivative of position v(t)=s'(t) They are using to find s'(1)

57. freckles

close

58. freckles

not exactly though

59. freckles

you are missing a sign

60. freckles

$v(1)=s'(1)=\lim_{h \rightarrow 0}\frac{s(1+h)-s(1)}{h}$

61. anonymous

-9?

62. freckles

yes

63. anonymous

My bad, didn't catch the negative sign.

64. freckles

are you using short cuts? or the long way?

65. freckles

because you found that really fast

66. anonymous

The long way but using mental math

67. freckles

nice

68. anonymous

I have the steps written down though, the long way. Would this be ok for an answer: First, I used: s(t) = -8 - 9t and t=1 to solve for (a+h) and (a). My 'a' term would be 1 in this case. f(a) = -8 - 9t = -8 - 9(1) = -17 f(a+h) = -8 - 9t = -8 - 9(1+h) = -17 - 9h Then I used the equation: f'(a) = lim[f(a+h) - f(a)/h] = f'(a) = -17 - 9h - (-17)/h = -9h/h = -9 Thus I found the derivative to be: f'(1) = -9

69. freckles

s(t) = -8 - 9t notice the slope of s(t) is -9 the derivative of s(t) is -9

70. freckles

but yeah your teacher probably wants you to use that formal definition of derivative thing but you know we remember from algebra that the slope of f(x)=mx+b is m and we learn from calculus that the slope is the derivative

71. anonymous

Ok, but is the way I showed my work using the formal definition?

72. freckles

yes it is pretty too

73. freckles

you did very well

74. freckles

only thing I would say and it is really not apart of your work "s(t) = -8 - 9t and t=1 to solve for (a+h) and (a). My 'a' term would be 1 in this case. :" I would say f(a+h) and f(a) but I knew what you meant when you said this

75. freckles

oops s(a+h) and s(a) since our function name is s not f

76. freckles

you could change all those f letters to s letters actually

77. anonymous

Ok thanks!

78. freckles

np

79. anonymous