## anonymous one year ago If the tangent line to y = f(x) at (4, 2) passes through the point (0, 1), find f(4) and f '(4). f(4)=2 , f'(4)=?

1. myininaya

so you can find the tangent line you are given two points and the slope use point-slope form of a line

2. anonymous

Y1-Y0=m(x1-xo)?

3. myininaya

actually you only need the slope

4. anonymous

yeah

5. myininaya

since that is what it asked for

6. myininaya

you know the slope formula right?

7. myininaya

$m=\frac{y_1-y_2}{x_1-x_2}$

8. anonymous

m=y1-y0/x1-xo

9. anonymous

is the other way around of how you put it

10. myininaya

1 and 0's 1 and 2's whatever

11. myininaya

do you know how to use the formula?

12. anonymous

is 1/4

13. myininaya

let's see top is 2-1 and bottom is 4-0 looks great

14. anonymous

yes

15. anonymous

do I use y=mx+b?

16. anonymous

b=2 if I do that with (4,2)

17. myininaya

you don't need to it just asked to find f'(4) and f(4) and f'(4) is the slope of the tangent line at (4,2) to the curve y=f(x)

18. myininaya

unless you want to find the tangent line ?

19. anonymous

I do

20. myininaya

it is just: $y-f(a)=f'(a)(x-a) \text{ is the tangent line \to the curve } y=f(x) \text{ at } (a,f(a))$

21. myininaya

you are given a is 4 here

22. myininaya

$y-f(4)=f'(4)(x-4)$

23. myininaya

now you found f(4) and f'(4)

24. anonymous

right

25. anonymous

thanks

26. myininaya

np

27. anonymous

problem the answer is not 2.

28. anonymous

I already tried that

29. myininaya

f(4) should be 2 it is given by (4,2) f'(4) should be 1/4 that is the slope of the tangent line at (4,2) to the curve y=f(x)

30. anonymous

oh yeah then I was right about the 1/4. in the question I put the f(4)

31. anonymous

perfect thank you

32. myininaya

yeah you already found f(4) earlier we were finding f'(4) and we used the slope formula because that is what the derivative means

33. anonymous

Find f '(a). f(x) = (1 − 2x)^1/2

34. myininaya

so do you know chain rule?

35. anonymous

yeah, saw this last year tho

36. anonymous

I thought it was 1/2(1-2x)^-1/2 * (2)

37. myininaya

$h(x)=f(g(x)) \\ h'(x)=g'(x) \cdot f'(g(x)) \\ \\ \text{ if } f(x)=x^n \\ \text{ then we have } \\ h(x)=(g(x))^n \\ h'(x)=g'(x) \cdot n(g(x))^{n-1}$

38. myininaya

well one complaint

39. myininaya

what is the derivative of 1-2x ?

40. myininaya

you put 2 but it should actually be...

41. anonymous

ooooh

42. anonymous

thanks

43. myininaya

but you can also simplify a bit

44. anonymous

hahaha hate when that happens

45. anonymous

yes i did before

46. anonymous

in my homework

47. myininaya

ok then you got it now :)

48. anonymous

do you have time for more?

49. anonymous

I had problems with this one too

50. myininaya

maybe one more then I have to go check on something

51. anonymous

(a) Find the slope m of the tangent to the curve y = 7/ x at the point where x = a > 0.

52. myininaya

is that just y=7/x?

53. anonymous

(b) Find equations of the tangent lines at the points (1, 7) and (4, 7/2) .

54. myininaya

$y=\frac{7}{x}=7 \cdot \frac{1}{x} =7 x^{-1} \text{ by law of exponents } \\ \text{ now differentiate } y=7 x^{-1} \text{ using power rule }$

55. anonymous

no x^1/2 sorry

56. myininaya

oh

57. anonymous

it didn't appear when i copy pasted it

58. anonymous

59. anonymous

but is not that

60. myininaya

$\text{ find the slope of the tangent line \to the curve } y=7x^\frac{1}{2} \text{ at } x=a>0$?

61. myininaya

or is that x^(1/2) in the denominator?

62. anonymous

ok part A i got it

63. anonymous

I had to change x=a

64. anonymous

now for B I don't now what I did wrong

65. myininaya

ok so it was: $y=\frac{7}{x^\frac{1}{2}}=7x^\frac{-1}{2}$

66. anonymous

I got that

67. anonymous

but I have part B wrong

68. myininaya

ok the tangent line at (1,7) is $y-f(a)=f'(a)(x-a) \\ \text{ here you have } a=1 \\ y-f(1)=f'(1)(x-1) \\ \text{ you already found } f'(x) \text{ \above } \\ \text{ you got that } f'(x)=\frac{-7}{2x^{\frac{3}{2}}} \text{ I think but I don't really know anymore } \\ \text{ \because I'm still trying \to figure out what } f(x) \text{ was :p } \\ \text{ so this becomes } \\ y-7=\frac{-7}{2(1)^\frac{3}{2}}(x-1)$

69. myininaya

and you still have to do this same process for the other point

70. anonymous

perfect

71. anonymous

is not that apparently