Author Topic: ANTIDIFF (Read 1796 times) Tweet Share

mandy · « **on:** August 17, 2009, 08:47:35 pm »

Find the equation of the curve whose gradient is described by f '(x)= x² - 2x - 6 and which passes through the point (1,3).

This is what I did:

Found the antidiff, then when I got the c at the end, I subbed in the y intercept (3).

This was wrongggg. The answer actually said the antidiff, with the c value as 29/3 :S

Why?

Flaming_Arrow · « **Reply #1 on:** August 17, 2009, 08:49:59 pm »

y intercept isnt 3

when you integrate

u get

y = x^3/3 - x^2 - 6x + c
sub in (1,3)
3 = 1/3 - 1 - 6 + c
c = 29/3

TrueTears · « **Reply #2 on:** August 17, 2009, 08:50:18 pm »

$f(x) = \frac{1}{3}x^3 - 2(\frac{1}{2})x^2 - 6x + c$

$(1,3) \to 3 = \frac{1}{3}(1)^3 - 2(\frac{1}{2})(1)^2 - 6(1) + c$

$\therefore \frac{-20}{3} + c = 3$

$\therefore c = \frac{29}{3}$

Thus $f(x) = \frac{1}{3}x^3 - x^2 - 6x + \frac{29}{3}$

EDIT: I was beaten by sailor mars.

mandy · « **Reply #3 on:** August 17, 2009, 08:55:16 pm »

Thanks, I understand now..

But, I'll probably be back with more qns, this exercise is driving me crazy.

monokekie · « **Reply #4 on:** August 17, 2009, 08:58:11 pm »

its driving me krazy atm too

mandy · « **Reply #5 on:** August 17, 2009, 09:20:58 pm »

How do I do this ?

Flaming_Arrow · « **Reply #6 on:** August 17, 2009, 09:27:18 pm »

divide first so u get 6x + 9 + 6/x^2

dcc · « **Reply #7 on:** August 17, 2009, 09:28:55 pm »

$ \begin{align*} \int \frac{3}{x^2} \left(2x^3 + 3x^2 + 2\right)\ dx &= 3 \int \left( \frac{2x^3 + 3x^2 + 2}{x^2} \right)\ dx\\ &= 3 \int \left(2x + 3 + 2x^{-2}\right)\ dx\\ &= 3 \left( \frac{2x^2}{2} + \frac{3x^1}{1} + \frac{2x^{-1}}{-1}\right) + C\\ \end{align*} $

Where we used $\int x^n \ dx = \frac{x^{n+1}}{n + 1} + C_i$ for all reasonable values of $n$ .

Gloamglozer · « **Reply #8 on:** August 18, 2009, 05:10:28 pm »

Quote from: TrueTears on August 17, 2009, 08:50:18 pm

$f(x) = \frac{1}{3}x^3 - 2(\frac{1}{2})x^2 - 6x + c$

$(3,1) \to 3 = \frac{1}{3}(1)^3 - 2(\frac{1}{2})(1)^2 - 6(1) + c$

With the last line in the above quote, you switched the values of x and y because the question said that the point (1,3) passed through the derivative and not the original function, right?

TrueTears · « **Reply #9 on:** August 18, 2009, 05:13:01 pm »

lol woops that was a typo i meant (1,3) the coordinates, thanks for picking that up XD

Gloamglozer · « **Reply #10 on:** August 18, 2009, 05:16:09 pm »

Your welcome.

But knowing you, I reckon you could re-write the rules and say it's correct. lol

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