from your
[2/(1-3x+h) - 2/(1-3x)]/h
it should have been
[2/(1-3(x+h) ) - 2/(1-3x)]/h
= [2(1-3x) - 2(1 - 3x - 3h)]/((1-3x-3h)(1-3x)) * 1/h
= [2 - 6x - 2 + 6x + 6h]/((1-3x-3h)(1-3x)) * 1/h
= 6h/((1-3x-3h)(1-3x) * 1/h
= 6/((1-3x-3h)(1-3x))
so the limit of that as h ---> 0 = 6/(1-3x)^2
so when x = 0, the slope = 6/(1-0)^2 = 6
and when x = 0 , in the original y = 1/(1-0) = 1
so the y-intercept would be 1 and the slope = 6
equation: y = 6x + 1
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.
1 answer