Help with Calculus Probelm please

[Kushballo7]

Alright so my calculus teacher likes to give us amazingly complex
problems without any hints on how to complete them. I really need help
so anything you have to offer will be great. The problems are as
follows.


1. Find the points P and Q on the parabola y = 1-x^2 so that triangle
ABC formed by the x-axis and the tangent lines at P and Q is an
equilateral triangle.


2. Find the two points on the curve y = x^4 - 2x^2 - x that have a
common tangent line.


3. If g(x) = f(b+mx) + f(b-mx) where f is differentiable at b, find
g'(0).


THANK YOU!!

 

[stapel]

1) By symmetry, whatever the x-value for Q on the negative x-axis is, the x-value for P on the positive x-axis will have the same absolute value. So let's say P is (p, 1 - p²) and Q is (-p, 1 - p²). And the third vertex of this triangle will have to lie on the y-axis.

On this curve y = 1 - x² we have y' = -2x = m at any point x. In particular, at P, the slope of the tangent is m = -2p, and the tangent-line equation is y - (1 - p²) = -2p(x - p). Then:

. . . . .y - 1 + p² = -2px + 2p²

. . . . .y = -2px + p² + 1

The x-intercept then will be at:

. . . . .0 = -2px + p² + 1

. . . . .x = (p² + 1) / (2p)

The y-intercept will be at:

. . . . .y = p² + 1

The distance between these points gives us the length of one of the sides. Since the side that lies on the x-axis is bisected by the y-axis, the length of a side is twice the distance of the x-intercept from the origin.

See what you can do with that.

Eliz.

 

[stapel]

3) Using the Chain Rule, we get:

. . . . .g'(x) = f'(b + mx)(m) + f'(b - mx)(m)

. . . . .g'(0) = f'(b)(m) + f'(b)(m) = 2mf'(b)

I don't see any way of going any further. Sorry.

Eliz.