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.

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

To find the equation of the tangent line to the curve 2/(1-3x) at x=0, you correctly applied the definition of the derivative. However, there seems to be a mistake in your algebraic manipulation. Let's go through the steps again to identify the error.

You started with the expression: [2/(1-3x+h) - 2/(1-3x)]/h.

To simplify this expression, we need to find a common denominator for the two fractions. The common denominator will be (1-3x)(1-3x+h), since the denominators are (1-3x) and (1-3x+h).

Now, let's continue with the calculation.

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

Simplifying the numerator:

[2 - 6x - 2 + 6x - 2h] / [(1-3x)(1-3x+h)] / h

Remove common terms:

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

Now, cancel h from the numerator and denominator:

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

Taking the limit as h approaches 0:

lim(h->0) -2 / [(1-3x)(1-3x+0)]

Now, simplify further:

-2 / (1-3x)^2

This is the expression for the derivative of the curve 2/(1-3x) with respect to x. To find the equation of the tangent line at x=0, we need to evaluate this derivative at x=0.

Substituting x=0 into the expression, we have:

-2 / (1-3(0))^2 = -2 / 1^2 = -2

So, the slope of the tangent line is -2.

To find the y-intercept (b) of the tangent line, we substitute the point (0, 2) (which lies on the original curve) into the equation y = mx + b, where m is the slope (-2):

2 = -2(0) + b
2 = b

Therefore, the y-intercept of the tangent line is 2.

Now we can write the equation of the tangent line:

y = mx + b
y = -2x + 2

So, the equation of the tangent line to the curve 2/(1-3x) at x=0 is y = -2x + 2, which is different from the answer y = 6x + 2 you provided.

The discrepancy might be due to the error that occurred during the algebraic manipulation. Please recheck your calculations and make sure to follow the steps mentioned here.