# Calculus

Let g and h be any two twice-differentiable functions that are defined for all real numbers and that satisfy the following properties for all x:
I) (g(x))^2 + (h(x))^2=1
ii) g'(x)= (h(x))^2
iii) h(x)>0
iv) g(0)=0

a)Justify that h'(x)=-g(x)h(x) for all x
b) Justify that ha has a relative maximum at x=0
c) Justify that the graph of g has a point of inflection at x=0

I was able to do a, but I couldn't figure out b or c.

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1. (a) using (i) and (ii)
2gg' + 2hh' = 0
2gh^2 + 2hh' = 0
gh + h' = 0
h' = -gh

(b) from (a) and (iv),
h'(0) = -g(0)*h(0) = 0
h" = -g'h-gh'
so,
h"(0) = -g'(0)h(0)
Since h(0)>0, g'(0)>0, so h"(0) < 0

(c) from (ii) and (a)
g" = 2hh' = 2h(-gh) = -2gh^2
g"(0) = 0 from (iv)

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posted by Steve
2. l) (g)^2 + (h)^2=1
so
2 g g' + 2 h h' = 0
g g' = -h h'
g' = -(h/g) h'

2)g' = h^2
and
h' = -g g'/h
h' = - g h^2/h = -gh that is part a

then
If h has a max, then h' has a zero
- g h = 0 ?
either g = 0 or h = 0
if h is max, look for zero of g
but we are given that g(0) = 0

c)
g' (0) = - g(0)h(0)
but g(0) = 0
so g'(0) = 0

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posted by Damon

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