(^308) Functions, graphs, and numbers
supposing that n > 1, show that the graph has a flexpoint at the point (xn,yn)
for which
cos xn = 1 or sin xn = 1 -
1\34
and ( 1)n12
n
n= 1-n
Remark: Unless we know about the famous number e, it is still not easy to esti-
mate yn when n is large. When we have learned the first of the formulas
lim 1 +xln= e, lim yn = e 3 = 0.606531,
n-- of
we will be able to put x = -1 and take square roots to obtain the second one.
(^11) This problem, like the preceding one, involves ideas. Supposing that
n > 1 and
(1) y = xn(2 - x)n = (2x - x2)",
show that
(2) dz= 2n(2n - 1)(2x - X2)-2rx2- 2x +
2n - 1
and hence that the graph of (1) has a flexpoint at the point (xn,yn) where
xn=1- Zn1 1' yn=[1-2n1 1]n.
12 Determination of the natures of the graphs of equations like
(1) y4 = x2(1 + x2)
is an ancient and honorable pastime. Observe that if the point (x,y) lies on the
graph, then so also do the points (x,-y), (-x,y), and (-x, -y). If we find the
partG of the graph in the first quadrant, we can therefore use symmetry to obtain
the rest of the graph. Henceforth we consider only points on the graph for
which x >- 0 and y >= 0. For these points,
(2) y = (x° + x2)3+.
To each x there corresponds exactly one y for which the point (x,y) is on G
Moreover, y > (x4)3a = x, so G lies on or above the line having the equation
y = x. Show that, when x > 0,
(3) dy 2x3 + x d2y x2(2x2 - 1)
dx 2(x4 + 7x2 4(x4 + x2)%i
Show that the slope is decreasing over the interval 0 < x < 1//, increasing
over the interval x > 1/', and attains the minimum value /62572s at the
flexpoint having the coordinates I/ N/'2- and Show that, when x > 0,
(4) 0 < `x+x2-x= 4 x2 < x2 2 - 1
(1/x4 + x2 + x)( x4 + x2 + x2) (2x )(2x) 4x