Derivatives: Composites, normal lines, n-th derivatives and more.

StopWatch

1. The problem statement, all variables and given/known data

1. The line perpendicular to the curve y = 2x^3 - x^2 + x - 3 at the point (1, -1) will intersect the x-axis at what point?

2. f(x) = |x^2 - 5| - x, for all x. Let g = f(f(f(x))), find g'(2). I tried just subbing in 2x - 1, the first derivative, to f(2x - 1) and then once more and ended up with 16 somehow, when the answer is -45.

3. If f(x) = ln(2X^2 + x - 1) - ln(x+1) find the 98th derivative at (1/2 + sqrt(2)/2). I know that the derivative simplifies to 2(2x - 1) and the 2nd derivative to -2/(2x-1)^2 and the 3rd to -8(2x-1)/2x-1)^4 but I always have a hard time generalizing these and then getting the answer (especially because the answer is -2^49(97!) and I have no idea how the factorial gets worked in.


2. Relevant equations

y1 - y0 = m(x1 - x0)



3. The attempt at a solution

Finding the derivative and subbing in x = 1 gives a slope of 5 at the point specified, which means m = -1/5. When solved this gives x = -10, however the correct answer is apparently -4.

This site is a godsend.

tiny-tim

Hi StopWatch!
Yes, but how did you then get -10 starting from x = 1, y = -1?

tiny-tim

But sometimes it's -2x - 1, isn't it?
Hint: what's the nth derivative of 1/x ?

StopWatch

Is the nth derivative (n!)x^-n? Does that make sense. Sorry for leaving for so long I had class. I hope someone's around though, the test is tomorrow morning and I have a ton of questions. I wish professors posted solutions to past tests. I redid the algebra and got 10 by the way y (which is zero when the x axis is intercepted) + 1 = 5(x - 1), so 0 = 5x -2 = 2/5 somehow I really screwed that up, wow.

It can be negative, but we're subbing in when x = 2 so those cases don't matter, or at least that's my logic. My main concern is about how to actually go about plugging into the composite like that.

I do have a new question as well though: The line perpendicular to x^3 - 2x + 1 at (2, 5) will intersect the x-axis at what point? I get (y-5) = 10(x-2) which gives me -15/10 for the x value of the tangent. I might just be really rushed in my thinking right now I'd really appreciate any help at all.

StopWatch

Anyone around?

tiny-tim

Hi StopWatch!

(just got up …)
yeees, but n-1 and times (-1)ⁿ
nooo, 5x - 6

but anyway it's (x - 1) = -5(1)
no, the slope of the tangent line is 10, so for the perpendicular line you need -10(y-5) = (x-2)

Anyway, good luck on your test this morning!