February 23, 2017

Homework Help: calculus

Posted by Daniel on Saturday, October 1, 2011 at 7:23pm.

I am trying to find equation of the tangent line to the curve 2/(1-3x) at x=0 using the definition.

I applied it into the formula
lim h->0 [f(a+h) - f(a)]/h

which gave me

[2/(1-3x+h) - 2/(1-3x)]/h

I took the conjugate and came with

[4/(1-3x+h)^2 - 4/(1-3x)^2] / h(2/(1-3x+h) + 2/(1-3x))

Then I expanded the top and I got

(-8h - 24xh - 4h^2) / [(1-3x+h)^2 (1-3x)^2] / h(2/(1-3x+h) + 2/(1-3x))

I pulled out the "h" in the numerator and cancelled the "h" on the top and bottom and got

(-8 - 24x - 4) / [(1-3x+h)^2 (1-3x)^2] / (2/(1-3x+h) + 2/(1-3x))

I applied the limit

(-12 - 24x ) / [(1-3x)^4] / (4/(1-3x))

I get

(-12 + 24x) / 4(1-3x)^3

From what I have done, it looks like my work is right, but the answer I come up with

y=-3x + 2

However, the answer is y=6x + 2

I can not see what I did wrong.

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