Verify my Calculus Please?
The question is asking:
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
Justify that h has a relative maximum at x=0
also;
Justify that the graph of g has a point of inflection at x=0.
For the first one... I have:
h'(x) = 0
h'(x)=-g(x)(h(x)) [this was the part a question]
-g(x)(h(x))=0
-g(x)=0 [now using rule IV:]
-g(0)=0; which means
h'(0) = 0- relative extrema. Any way to prove this is a MAXIMUM instead of just extrema?
For my second question:
g'(x)=(h(x))^2
g''(x)=2(h(x))(h'(x))
g''(x)=0 [because h'(0) = 0]
so g''(0)=0 is a point of inflection when x = 0.
Is there any way to prove that besides having it = zero? Maybe using h(x)>0 in some way?
My questions are just asking about proof of the answer, if anyone can help me I'd be really appreciative.
Comments
You are doing just fine!
To prove that x = 0 is a maximum for h and
point of inflection for g use this information:
(ii) g'(x) = (h(x))^2
(iii) h(x) > 0
(iv) g(0)=0
h(x) is always positive (non zero !) so g'(x) = (h(x))^2 >0
which means that the function g(x) is strictly increasing.
As g(0) = 0 and g(x) increases so g(x) <0 when x<0
and g(x)>0 when x>0.
With this information you can prove that h'(x) changes sign
from - to + going through x = 0 ==> it is max
The same with g"(x).
Hope, it will help