Question

Find the particular solution of the differential equation f''(x)=6(x-1) whose graph passes through (2,1) and is tangent to the line 3x-y-5=0 at that point

Answer 1

i know how to find the integral but im wondernig about the first C value

Answer 2

Would you like me to teach you @rjeffries96 ?

Answer 3

@calculusfunctions SHE REALLY NEEDS HELP AND SHE HASTO GET OFF AT 630 sorry for the caps

Answer 4

im just confused about how to find the first c by using the tangent line portion of the problem (thanks mikala1)

Answer 5

Alright but @mikala1 you know me well enough now to know that I'm not simply going to just give her the answer? BTW Hi!

Answer 6

haha hi :) so am i right when i say that the first derivative and the equation given as the tangent line will be the same line?

Answer 7

@rjeffries96 no problem can you write down the answer you have thus far so that I can see that you've done it correctly?

Answer 8

well right now i have that the integral of 6x-6 = 3x^2 -6x +C

Answer 9

i need help finding the C value, then i know to take the 2nd integral and sub in the point given to yield f(x)

Answer 10

Now you know that f '(x) is the slope. You also know that since the tangent to this curve is

3x - y - 5 = 0

The slope of this tangent equals f '(x).

Answer 11

Ok, that makes sense. But then how would I find C? would i set the two equations equal (I tried and it wasnt much use) or do i substitute in the values x=2 and y=1?

Answer 12

No the point (2, 1) is for f(x). All you know at the moment is that f '(x) = 3. Before I proceed further, are you sure you wrote the entire question, as is? Double check please?

Answer 13

Yes, that is the entire question

Answer 14

how did you find that f'(x)=3?

Answer 15

Because the slope of the given tangent line is 3.

Answer 16

oh ok (oops i shoulve seen that)

Answer 17

its 6 30 sorry to interupt

Answer 18

im really sorry but i have to be at school in 10 minutes, i can just ask my teacher. thanks anyways

Answer 19

well did he help you in any way

Answer 20

OK when you're back on next time, let me know and I'll show you if you still don't understand.

Answer 21

thanks for helping him/her lol she/him told me to tell u but she was late ololo

Answer 22

It's Ok @mikala1 I'll break the rules just this once and post the solution for her.

Answer 23

i think she got off but when ever she gets on

Answer 24

f ''(x) = 6x - 6

f '(x) = 3x² - 6x + C

We are told that 3x - y - 5 = 0 is tangent to the curve at the point (2, 1). This implies that both the curve and the tangent line pass through this point. Hence the first derivative (slope) is 3 at x = 2.

Thus if f '(x) = 3x² - 6x + C then
3 = 3(2)² - 6(2) + C
3 = 12 - 12 + C
C = 3

Thus f '(x) = 3x² - 6x + 3 Now

f(x) = x³ - 3x² + 3x + K

Now f(x) passes through (2, 1). Thus

1 = (2)³ - 3(2)² + 3(2) + K
1 = 8 - 12 + 6 + K
K = -1

∴ f(x) = x³ - 3x² + 3x - 1

Answer 25

@mikala1 if for some reason you're able to find her right now, please ask her to check it out.

Answer 26

ok she left for school so she will be on when she gets home she said but i think it is great and easy to understand

Answer 27

Thank you @mikala1 I have to sign out now but before I do, do you need anything?

Answer 28

no thank you byyyy

Answer 29

BYE!!!