G(x) in respect to F(X)

**Rimas** · March 17th 2009, 03:41 PM

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)Prove that $h'(x)=-g(x)h(x)$ for all x
(b)Prove that h has a relative maximum at x=o
(c)Prove that the graph of g has a point of inflection at x=o

**skeeter** · March 17th 2009, 04:49 PM

Originally Posted by Rimas

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)Prove that $h'(x)=-g(x)h(x)$ for all x
(b)Prove that h has a relative maximum at x=o
(c)Prove that the graph of g has a point of inflection at x=o

(a) using the chain rule, take the derivative of equation I

use equation II for a substitution into the derivative equation ... prove (a).

(b) use the result from (a) to find h'(x) and h''(x) ... show h'(0) = 0 and h''(0) < 0

(c) find g''(x) from equation II and show that g''(0) = 0 and that g'' changes signs at x = 0.

**OnMyWayToBeAMathProffesor** · March 18th 2009, 07:00 PM

How do you prove that $h''(0)<0$ and why? I understand that when the second derivative is below the x-axis the slop is negative and it is concave down. Also it means that the slope of the first derivative is negative. How would all this help?

and how would you prove $g''(x)$ changes sign at $x=0$ ?

anybody?

**Jhevon** · March 18th 2009, 07:15 PM

Originally Posted by OnMyWayToBeAMathProffesor

How do you prove that $h''(0)<0$

differentiate h'(x) = -g(x)h(x) and plug in x = 0

and why?

we are asked to prove that h(x) has a relative maximum at x = 0. for this, we need to know that h'(0) = 0 and we want to find h''(0) to use the second derivative test

I understand that when the second derivative is below the x-axis the slop is negative and it is concave down. Also it means that the slope of the first derivative is negative. How would all this help?

by the second derivative test, h'(0) = 0 and h''(0) < 0 implies that x = 0 yields a relative maximum. h''(0) < 0 means that h(x) is concave down about zero. this shows it is a maximum since at zero we have a critical point.

and how would you prove $g''(x)$ changes sign at $x=0$ ?

it suffices to show that g''(0) = 0 and g'(0) does NOT change signs. (why?)

**skeeter** · March 18th 2009, 07:23 PM

Originally Posted by OnMyWayToBeAMathProffesor

How do you prove that $h''(0)<0$ and why? I understand that when the second derivative is below the x-axis the slop is negative and it is concave down. Also it means that the slope of the first derivative is negative. How would all this help?

and how would you prove $g''(x)$ changes sign at $x=0$ ?

anybody?

$g'(x) = [h(x)]^2$

$g''(x) = 2h(x) \cdot h'(x)$

since $h'(x) = -g(x) \cdot h(x)$ ...

$g''(x) = 2h(x) \cdot -g(x) \cdot h(x) = -2g(x)[h(x)]^2$

note that $[h(x)]^2 > 0$ for all $x$

given $g(0) = 0$ and $g'(x) = [h(x)]^2 > 0$ ... $g(x)$ is always increasing

since $g(x)$ is twice differentiable, what does that tell you about the value of $g(x)$ for $x < 0$ and $x > 0$ ?

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