Primitive Functions Problem (1 Viewer)

[Kulazzi]

Hey guys!

I'm just (trying) to finish off an exercise that we were supposed to do before the holidays came. Well, I tackled it this morning and I have 6 problems that I cannot seem to solve from this exercise. I'll post them up, type up what I tried to do to solve them, and please, anyone point out any silly mistakes or any steps I may have missed, or if I have the whole working out process wrong, can someone please type up the right procedure. Right, thanks, here we go........oh yeah, as the heading says, it's primitive functions

1) Express y in terms of x if:y' = 2/x^2

y' = 2 / x^2

y = 2x / [x^3/3] +C

= 2x + 3 / x^3 + C

Answer at back of book: -2 / x + C (No working out shown)

2) If f'(x) = 4x - 7 and f(2) = 5, find f(x)

f'(x) = 4x - 7

f(x) = 2x^2 - 7x + C

sub in f(2) to find the value of C

f(2)....ok I just figured out the answer. I forgot to put the whole equation equal to 5 [f(2) = 5]. See? sometimes it works when you tell someone else about it. Moving on.

3) Given f'(x) = 3x^2 + 4x - 2 and f(-3) = 4, find f(1)

same situation as above. Forgot that f(-3) = 4, therfore forgot to put whole eq equal to 4, to workout C. doh!

4) Given y'' = 8 and y' = 0 and y = 3 when x = 1, find the equation of y in terms of x

y'' = 8

y' = 0

when x = 1, y = 3

I had no idea where to go from here.

Answer at back of book: y = 4x^2 - 8x +7 (No working out shown)

5) If y'' = 12x + 6 and y' = 1 at the point (-1, -2), find the equation of the curve.

y'' = 12x + 6

y' = 12 [x^2 / 2] + 6x + C

= 6x^2 + 6x + C

y' = 1

1 = 6x^2 + 6x +C

sub in x = -1

1 = 6 - 6 +C

therefore, C = 1

y' = 6x^2 +6x +1

y = 2x^3 + 3x^2 + x +C --> Stopped here.

Answer at back of book: y = 2x^3 + 3x^2 + x - 2 (No working out shown)

6) If f"(x) = 6x - 2 and f'(2) = f(2) = 7, find the function f(x)

f"(x) = 6x - 2

f'(x) = 3x^2 - 2x + C

f'(2) = 7

7 = 12 - 4 + C

7 = 8 + C

C = -1

f'(x) = 3x^2 - 2x - 1 ---> From here, I figured out what I did wrong. I didn't intregrate properly into the normal eq --> f(x) = x^3 - x^2 - x + C, what I had written down was f(x) = 3x^2 - 2x - x + C

Any help would be greatly appreciated. Thanx again guys.

 

[...]

1) Express y in terms of x if:y' = 2/x^2

y' = 2 / x^2

y = 2x / [x^3/3] +C

= 2x + 3 / x^3 + C

take 2 out and integrate it as 1/x^2

for question 4), i think u can approach it like u did for 2) & 3)

for 5) you just sub in the values of (-1, 2) into ur final equation to get C

 

[Linda N]

1) y' = 2/x^2

y = 2 . 1/x^2
= 2 . Integral of [x^-2]
= 2 . [x^-1/ -1] + C

:. y = -2/x + c

2) f '(x) = 4x - 7
f(x) = 2x^2 - 7x +c

subst. x=2 , f(x) = 5

5 = 2(2)^2 - 7(2) +c
5 = 8 - 14 +c
c = 11

:. f(x) = 2x^2 - 7x +11

3) f'(x) = 3x^2 + 4x - 2
f(x) = x^3 + 2x^2 - 2x +c

subst. f(-3) = 4

4 = (-3)^3 + 2(-3)^2 - 2(-3) +c
4 = -27 + 18 +6 +c
c = 7

:. f(x) = x^3 + 2x^2 - 2x +7

subst. f(1)

f(1) = 1^3 + 2(1)^2 - 2(1) +7
= 8

4) y" = 8
y' = 8x +c

subst. x=1, y' = 0

0 = 8 + c
c = - 8

:. y' = 8x - 8
y = 4x^2 - 8x + c

subst. x=1, y=3

3 = 4(1)^2 - 8 +c
c = 7

:. y = 4x^2 - 8x + 7

 

[Kulazzi]

thanx ... and Linda for your help