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.
(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)
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
b) Well, well, it's time for some maximum humor! To justify that h(a) has a relative maximum at x=0, let's take a closer look at our functions. We know that g(0) = 0, which means that at the point x=0, our clownish friend g is taking a little break from being a non-stop joker. Since g'(x) = (h(x))^2, we can deduce that g'(0) = (h(0))^2. Let's amplify the hilarity by differentiating both sides of equation I): [2g(x)g'(x) + 2h(x)h'(x)] = 0. Remember, we're at x=0, so this equation summing up the comedic duo becomes [2g(0)g'(0) + 2h(0)h'(0)] = 0, and since g(0) = 0, we can simplify to [2h(0)h'(0)] = 0. It seems like we've stumbled upon a potential set-up for the punchline! Since h(0) > 0 (property iii), we can conclude that h'(0) = 0. So voila! h(a) has a relative maximum at x=0, where our clownish function partner g is taking a little nap.
c) Brace yourself for a hilarious point of inflection! To justify that the graph of g has a point of inflection at x=0, let's analyze the second derivative of our clownish friend g. We know that g'(x) = (h(x))^2, and we've also figured out from part b) that g'(0) = 0. Now, let's differentiate g'(x) = (h(x))^2 with respect to x to find the comedic twist (our second derivative): g''(x) = 2h(x)h'(x). Plug x=0 into this equation, and you get g''(0) = 2h(0)h'(0) = 0. Ah, the comic timing is impeccable! Since h(0) > 0 and h'(0) = 0 (from part b)), we deduce that g''(0) = 0. Thus, g has a point of inflection at x=0, where the humor takes a sudden turn.
To justify parts b) and c), we need to analyze the properties of the functions and their derivatives.
b) Justification for a relative maximum of h(x) at x=0:
To determine if h(x) has a relative maximum at x=0, we can use the first derivative test. According to the given properties, we know that g'(x) = (h(x))^2, and we need to find h'(x).
To calculate h'(x), we can differentiate the equation given in property I):
(d/dx) [(g(x))^2 + (h(x))^2] = (d/dx) [1]
2g(x)g'(x) + 2h(x)h'(x) = 0
Since g'(x) = (h(x))^2, we can substitute this in the equation:
2g(x)(h(x))^2 + 2h(x)h'(x) = 0
From property iv), we know that g(0) = 0, so at x=0, g(x) = 0. Substituting this in the equation gives:
2(0)(h(x))^2 + 2h(x)h'(x) = 0
2h(x)h'(x) = 0
To find the critical point(s), where h'(x) = 0, we can divide both sides by 2h(x):
h'(x) = 0, for h(x) ≠ 0
Since property iii) states that h(x) > 0, there are no values where h(x) equals zero. Therefore, h'(x) is never equal to zero for any x (except possibly at x=0).
Based on the first derivative test, if h'(x) is always positive for x less than 0 and always negative for x greater than 0, then h(x) has a relative maximum at x=0.
To analyze h'(x), we need to use the equation h'(x) = -g(x)h(x), which we derived in part a). Since g(x) = 0 at x=0, h'(x) = -g(x)h(x) = 0 at x=0. From this, we can deduce that h'(x) is positive for x < 0 and negative for x > 0.
Therefore, h(x) has a relative maximum at x=0.
c) Justification for a point of inflection of g(x) at x=0:
For a point to be an inflection point, the concavity of the function must change. To determine if g(x) has a point of inflection at x=0, we need to analyze the second derivative, g''(x).
Differentiating g'(x) = (h(x))^2 with respect to x gives us:
g''(x) = 2h(x)h'(x)
Substituting h'(x) = -g(x)h(x) (derived in part a)), we get:
g''(x) = 2h(x)(-g(x)h(x))
g''(x) = -2g(x)(h(x))^2
At x=0, we can evaluate g''(x):
g''(0) = -2g(0)(h(0))^2
Since g(0) = 0 (property iv), g''(0) = 0 as well.
If g''(x) changes sign around x=0, we can conclude that g(x) has a point of inflection at x=0. Let's analyze g''(x) using the properties we know:
Since we have h(x) > 0 (property iii), at all points, (h(x))^2 is always positive.
From property iv), g(0) = 0, and since g(x) = 0 around x=0, the factor g(x) is either positive or negative but doesn't change sign.
Therefore, g''(x) = -2g(x)(h(x))^2 is always negative or always positive for any x. Thus, there is no change in concavity, and the graph of g(x) does not have a point of inflection at x=0.